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Could one theoretically use the expansion of the universe to travel through it? At least in one direction? Could one theoretically use the expansion of the universe to travel through it? At least In one direction?
That’s it that’s my question.
I’m not a physicist but I do get ideas.
I also wonder if one could theoretically ride the magnetic fields in the universe and if space is like water could it be possible to “slap it and ride the wave”?
| *
*Yes, you are traveling the universe right now due to it's expansion, at a relative speed of 67.4 km/s per megaparsec, which is quite fast. Unfortunately there is no easy way you can change direction or speed, so you can just enjoy the ride.
*You can utilize magnetic fields to change your speed, but magnetic fields found in interstellar space are not strong enough to be of much use. Magnetic fields are currently used to slowly rotate satellites (or more precisely to unload control moment gyroscopes).
*"Slap and ride" is not feasible. There is nothing you can easily slap. Space is just too empty. Yes, even in empty space there is dark matter, neutrinos - but they don't like to interact with anything.
| {
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Why are solar panels kept tilted? I have noticed that, in my country India, most of the solar panels are tilted southward at an angle of ${45}^{\circ} .$ Even on buildings with inverted V-shaped roofs, solar panels are still oriented southward on both the sides of roof.
Research
Many sites suggests that the tilt aids in self-cleaning also another site stated that tilt depends on factor like latitude
My questions:
*
*Why are solar panels tilted southward?
*How is latitude of the location of a solar panel relevant in increasing efficiency?
| Its because the sun is never quite in the same part of the sky during the year. Different tilts will generate more energy at different times of year - eg a report for the island of Sark gave 2 tilt options:
So 60 degree tilt gives more in winter, but less in summer, with a small loss of total efficiency. No doubt India is different for solar generation, but that could be why panels are tilted like they are.
| {
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How did Maxwell figure out the speed of light? The Wiki article is about 2 graduate years of physics beyond my understanding. What is a good high-school rendition of his thought process: regarding his use of the "distributed capacitance and inductance of the vacuum" to reach his conclusion?"
| Maxwell derived that the speed of propagation of electromagnetic waves was
$$c = \sqrt \frac{1}{\epsilon_0 \mu_0},$$
where epsilon and mu were known from experiment.
His result for $c$ was close to the experimental value of the speed of light in his days, so he could conclude that also light was an electromagnetic wave. (For better history, there is https://hsm.stackexchange.com/ .)
| {
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Momentum anti-Hermitian in RQM? So in any book that discusses the Dirac equation in Relativistic Quantum Mechanics (For example you can check Bjorken and Drell, Greiner, or Ashok Das), whenever they take the dagger of the Dirac eq., they replace $i \frac{\partial}{\partial x}$ by $-i \frac{\partial}{\partial x}$, which is the same as $\hat{p} \rightarrow \hat{p}^\dagger=-\hat{p}$. In quantum mechanics, however, we found that $\frac{\partial}{\partial x}^\dagger=-\frac{\partial}{\partial x}$ which came from doing integration by parts and the vanishing of the boundary term. How do we reconcile both views and is $\hat{p}$ really anti-Hermitian in RQM?
| $\hat{p}^\dagger = -\hat{p}$ is not correct. Momentum operator churns out the momentum of a particle/system which is a real observable. Hence, it has to be Hermitian, since the expectation value of any anti-Hermitian operator must be purely imaginary and hence can't be observed.
How do we see momentum is Hermitian:
*
*$\dfrac{d}{dx}$ is anti-Hermitian: $\left(\dfrac{d}{dx}\right)^\dagger = -\dfrac{d}{dx}$.
*Since $\dfrac{d}{dx}$ is anti-Hermitian, $\left(i\dfrac{d}{dx}\right)$ must be Hermitian: $\left(i\dfrac{d}{dx}\right)^\dagger= - i\left(-\dfrac{d}{dx}\right) = \left(i\dfrac{d}{dx}\right)$.
*From this relation the momentum operator, $\hat{p} = - i \hbar\dfrac{d}{dx}$ is indeed Hermitian: $\hat{p}^\dagger=\hat{p}$.
I guess you are trying to say this probably: $\hat{p}^* = \left(-i \hbar\dfrac{d}{dx}\right)^* = \left(i\hbar\dfrac{d}{dx}\right) = -\hat{p}$.
| {
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Factorising a 4D Dirac delta function in a rest frame I'm working through a QFT problem and at one stage in the solutions we have this step:
$$\delta^{(4)}(p - q_1 - q_2) = \delta(E_1 +E_2 - M)\delta^{(3)}(\bf{q_1} - \bf{q_2}).$$
We are working in the rest frame of a meson with mass $M$ and the process is a decay to a nucleon anti-nucleon pair.
I cannot quite see why we are allowed to split the delta function this way. Can anyone break this down further for me?
| We can quickly show this using a limit definition of the dirac delta, that is
\begin{eqnarray*}
\delta(x_1,\dots,x_n) &=& \lim_{\epsilon \to 0^+}\frac{1}{\epsilon^n}e^{-\pi (x_1^2+\dots + x_n^2) / \epsilon^2} \\
&=& \lim_{\epsilon \to 0^+}\frac{1}{\epsilon^n}e^{-\pi x_1^2 / \epsilon^2}\times\dots \times e^{-\pi x_n^2 / \epsilon^2} \\
&=& \delta(x_1)\times \dots \times \delta(x_n)
\end{eqnarray*}
All you need to do is recognize that
$$ p - q_1 - q_2 = (p^0-q^0_1-q^0_2, \vec{p}-\vec{q}_1- \vec{q}_2)$$
You can put this straight into the definition above and get the required pieces (for example if $p$ represents the meson four momentum, clearly it only has one component which is $p^0 = M$ and $\vec{p} = \vec{0}$)
| {
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What will happen to atmospheres of very large planets? Our earth can hold our atmosphere whereas Mars cannot. So the atmosphere retention mass must be between the masses of Mars and Earth, but if mass is to be considered then can an iron ball having the same mass of the earth hold an atmosphere of its own?
Also since larger planets can hold larger atmospheres, if a planet like Jupiter was like the Earth in composition and the atmosphere contained gases in the same ratio as here on earth, will it result in the atmosphere being modified to suit the conditions of higher pressure and gravity of the planet?
What kind of atmospheric changes could be expected?
Can a very large planet result in the liquefaction of gases (I know that planet size is not the prerequisite for the presence of liquified gases but), if it does, can it be due to gravity alone?
| It must be enough to curve space-time around it significantly.But the more mass an object has doesnt mean the more atmosphere it will have.Neutron stars , which have 3 or 4 times the mass of the size have an atmosphere of 10 centimeters maximum . See :Neutron Stars - Kurzgesagt
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Why do Candelas and Howard say that $\sum_{n=1}^\infty \cos\left( n \kappa \epsilon \right) \ = \ - \frac{1}{2}$? In the paper Vacuum $\langle \phi^2 \rangle$ in Schwarzschild Spacetime by Candalas and Howard, they say that for each non-zero $\epsilon$ it is true that
$$
\sum_{n=1}^\infty \cos\left( n \kappa \epsilon \right) \ = \ - \frac{1}{2}
$$
This is equation (2.7) in the paper, where $\kappa$ is a constant (later set as the surface gravity for the black hole) and $\epsilon \to 0^{+}$ is taken as a regulator.
In what sense is this true? As some kind of distributional statement? Because this doesn't converge in the strict sense.
| Note that
$$III(x)~=~ \delta(x-\mathbb{Z})~=~ \sum_{m\in\mathbb{Z}} \delta(x-m)~=~\sum_{n\in\mathbb{Z}}e^{2\pi i xn}~=~1+2\sum_{n\in\mathbb{N}}\cos(2\pi xn)$$
is the Dirac comb/Shah distribution.
| {
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Why can't a low solidity wind turbine be used in high torque applications through gearing? From my understanding, low solidity wind turbines (such as the three blade type) are more efficient due to a higher tip speed ratio, giving a higher coefficient of performance. However, they are not well suited to high torque applications.
Whereas, high solidity turbines have a lower tip speed ratio hence they are less efficient, but they produce more torque, making them better suited for applications like pumping water.
This is what the text books tell me: low solidity for electricity production vs high solidity for pumping water.
My question; why not just use high speed, low solidity turbines and gear them to produce a low speed, high torque output? This makes use of the more efficient turbine design and allows it to be applied to a high torque requirement. There are other factors in choosing a turbine, such as start up speeds but surely in terms of power output gearing can be used to match a turbine to a load?
Thanks!
| Wind turbines are application-designed, and adding gears (to address different applications) costs money. In addition, a many-bladed turbine is efficient enough as-is to run a water pump given that the required work comes for free. You just scale up the blade disc diameter a bit if you want more work output and live with the inefficiency.
| {
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Pressure in irrigation water pipe I would like to know the pressure at the bottom of the pipe which I use for irrigation purpose. The land is step cultivated and the pipe goes slanting for the length of 140 meters and the top height would be 45 meters, water is pumped out from borewell using 7.5 hp submersible pump. I would like to know the pressure at the bottom of the pipe (Surface of the land, borewell depth can be ignored) when the water reaches the top.
| It will be the pressure that you find at 45 meters depth in water, that is 543 kPa or equivalently 5.28 atm (my reference).
I think that when the flow of water in the pipe changes rapidly this value will change.
The result is a consequence of the fact that the gradient of pressure into a static liquid is proportional to the gravitational field (or the gradient of the total potential).
| {
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Which side is convex in my plano convex lens? So, I am building an optical system and have got a Thorlabs plano-convex lens (part # LA1172-C) with a 400mm focal length. This makes the convex side of the lens so flat that it is difficult for me to discern (by naked eye) which side is convex on the lens. I need to know which side is convex in order to build my optical system.
Is there a good technique to figure out the convex side of my plano convex lens?
| In general in thorlabs lens there is a little > pointing the curved surface or written with a pencil or printed in the side of the lens.
If not, in general people place the lens under a long rectilign neon lamp and look at the reflexion on both surfaces. Is the neon reflexion seem curved on one of the surface? This is the curved surface.
With a 400mm focal lenght it is feasible, with a focal lenght higher than 1000mm it is becoming more complicate to see the difference.
| {
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Questions about BRST symmetry For a course about the standard model, I am writing a paper on BRST symmetry. For this I am mainly following the material developed in chapter 16.4 of Peskin and Schroeder. I am mostly done, however there are still two questions remaining that I cannot seem to answer.
*
*On page 518, they write that the second BRST variation of the ghost field is given by
\begin{equation}
Q^2 c^a = \frac{1}{2}g^2 f^{abc}f^{bde}c^cc^dc^e. \tag{16.49}
\end{equation}
I am able to prove this fact but then they claim that this vanishes due to the Jacobi identity for the structure constants $f^{abc}$. Here, I have tried to prove this but I can never show this. Does anybody have any hint how you would prove this?
*On page 519 they introduce the conserved charge $Q$ associated to the BRST symmetry. On the same page they seem to use the fact that this operator is Hermitian. (To prove that states in $\mathcal{H}_2={\rm Im}(Q)$ have zero inner product with each other.) However, I see no reason why this operator should in fact be Hermitian. Can anyone enlighten me as to why this fact is true?
| *
*Nilpotency follows from the fact that $c^cc^dc^e$ is cyclic in the indices $c$, $d$ & $e$, so that we may replace the structure constants $ff$' in the formula for $Q^2$ with the full Jacobi identity divided by 3.
*The Hermiticity of the BRST charge $Q$ is a first principle of BRST theory. In Yang-Mills theory, it basically follows from the Hermiticity (possibly anti-Hermiticity in some conventions) of the fundamental variables of the theory.
| {
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Liouville's integrability theorem: action-angle variables For classical dynamical systems, let $I_{\alpha}$ stand for independent constants of motion which commute with each other. 'Remark 11.12' on pg 443 of Fasano-Marmi's 'Analytical Mechanics' suggest that $I_{\alpha}$s can be taken as canonical coordinates.
For a conservative system, the Hamiltonian $H$ is a constant of motion. Let's refer to $H$ as $I_1$. Then $I_1$ becomes one of the canonical momenta. Hence $H$ can be written as $H=I_1$. Application of Hamilton's eqns. of motion implies that only one angle variable $\phi_1$ (corresponding to $I_1$) evolves linearly in time while all others stay constant because
$$
\dot{\phi_i}=\frac{\partial H}{\partial I_i} = 0 ~~~~~~~~~~~~~~~~~\mathrm{for~}i\neq1.
$$
So, is it true that for every Liouville integrable (described here) and conservative system (where Hamiltonian does not depend on time explicitly), Hamiltonian can be written as a function of only one action variable $I_1$ and only one angle variable (corresponding to $I_1$) evolves linearly in time, whereas others stay constant?
| In the Hamiton-Jacobi (HJ) approach, the Hamiltonian does not stay the same. It changes via (Eq. 9.17-c of Goldstein)
$$
K = H + \frac{\partial F_2}{\partial t},
$$
where $K$ is the transformed Hamiltonian. In HJ approach, we tune $F_2$ in such a way that $K=0$ (Eq. 10.2 of Goldstein). The above question assumes that $K=H$ which is possible only if $F_2$ is independent of $t$ which is not what happens in the HJ theory.
| {
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Binding energy and strong force If binding energy is responsible for holding nucleons together than what is meaning of strong force?
| Whenever there is a force, of any kind, there can be energy associated with what the force does. For example, if the force is attractive then it can hold two things together. In this case to pull the things apart you would have to apply a counter-force, and as the things moved apart, pulled by such a counter-force, energy is being supplied. The total amount of energy that would be required is given by multiplying the force at each stage by the next small amount of distance moved by the body the force is acting on, and then adding up all these contributions. The total is called binding energy.
In the case of nucleons the force is the one called "strong nuclear force" and the energy required to pull the nucleons apart is called their binding energy. These are two aspects of the same physical effect.
There are examples in other areas. For example, molecules are held together by electromagnetic forces, and the energy required to separate them is again called binding energy.
| {
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Does light take the path of least time because it travels in straight lines or vice versa? My question is which of these two feats is a consequence of the other?
Light travels in straight lines, mostly. Does it do that as a result of Fermat's principle of least time? and if so, is there a reason as to why it follows the path of least time? or is this another "that's the way the universe works" question?
And by reason I mean a physical explanation not mathematical deduction.
Or is it the other way around? meaning light taking the path of least time is just an obvious manifestation of the fact that it goes in straight lines?
| Feynman used to say that a good physicist knows many different ways to arrive at the same result. If you assume Fermat's Principle, you will get that light travels at straight line. If you assume light travels in straight lines and snell's law, you will get Fermat's Principle. This equivalence makes it hard to decide which one is the "fundamental" description.
I personally think that the Differential Equation is always the correct form to consider as the fundamental one. A differential equation describe the evolution of a state through time, which is what one observes. This view also lets you avoid some relativity problems.
So to your question, the answer will be that both results are direct byproducts of Maxwell's Equations, making neither one of them more fundamental than the other.
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What is the wavelength at the classical turning points using WKB Approximation? According to what I know is that a classical turning point in Newtonian Mechanics is a point where a particle has a zero kinetic energy (Total energy is equal to potential energy) and must be instantaneously at rest. This means it stop its motion and reverse direction similar to harmonic motion oscillating back and forth between points $x=-A$ and $x=+A$.
In the equation given below doesn't the wavelength
$$ \lambda(x) = \frac{h}{\sqrt{2m(E-U(x))}} $$
tend to infinity when $E=U(x)$?
| *
*The WKB approximation is valid in regions where
$$\left| \frac{d\lambda(x)}{dx} \right|~\ll~2\pi, \tag{46.6}$$
i.e. the WKB approximation breaks down near a turning point $x$, cf. Ref. 1. Generically, the velocity/momentum behaves as $\sim |x-x_0|^{1/2}$, so that the de Broglie wavelength has a singularity $\sim |x-x_0|^{-1/2}$
*Recall that the wavefunction solution $\psi$ to the 2nd-order TISE has 2 integration constants. Typically the solutions in two semiclassical regions [satisfying (46.6)] on each side of a turning point are matched via WKB connection formulas.
References:
*
*[LL] L.D. Landau & E.M. Lifshitz, QM, Vol. 3, 2nd & 3rd ed, 1981; $\S46$.
| {
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Why does thermodynamics use the negative of the Legendre transform? So I see how the negative Legendre transform is very helpful in interchanging dependencies and giving us the four different major thermodynamic potentials, from internal energy to Helmholtz, Gibbs, and enthalpy.
But what I'm unclear on is why we use the negative Legendre transform. Is it more that people derived the four above thermodynamic potentials by physical arguments, and then noticed that you have to use the negative Legendre transform to move between them, or is there some principled physical reason for why the negative Legendre transform makes more physical sense than the usual Legendre transform?
Any thoughts appreciated.
| The choice of the sign of the Legendre transform is purely conventional. There is no obliged way of defining it. The mathematical definition, used in physics in the case of the passage from Lagrangian to Hamiltonian function, has some advantages with respect to the opposite sign convention used in thermodynamics. Probably one of the most important is the conservation of the kind of convexity: if $f(x)$ is a convex function, its Legendre transform $\phi(p)=\sup_x (xp-f(x))$ remains a convex function, while with the opposite choice for the sign it would become concave.
However, permanence of convexity is just one of the possible reasons for a choice. Another, which probably played a major role in Thermodynamics, although I am not aware of explicit mentioning of such a point in books or papers, is to preserve an energy-like character of the other thermodynamic potentials. By energy-like character I mean that a positive heat transfer to the system, or a positive work done on the system would result in an increase of the corresponding thermodynamic potential.. For example, if enthalpy would had been defined as
$$
\tilde H(P,S) = -PV -U
$$
its differential would be
$$
d \tilde H = -VdP -TdS
$$
with the consequence that adding positive heat $TdS$ to the system, at constant pressure, would decrease such a redefined enthalpy.
Of course, nothing would be wrong with such an alternative choice, and the physics would remain exactly the same.
| {
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Thermodynamic description of few-body systems How large should a system be to become thermal?
Thermodynamic description is well-established for systems with large numbers (say, of order of $N_A\sim 10^{23}$) of constituents. Is there a "lower bound" of sorts, for the number of degrees of freedom $N$ in a system, for which thermodynamic notions such as temperature, entropy, etc., still remain applicable?
Thank you!
| Check this link out:
Statistical mechanics of Henon-Heiles oscillator
It is shown that although the system has only 2 degrees of freedom, due to the non-linearity the system exhibits ergodic character, which is endemic of systems with large DoF.
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Origin of the energy distribution of synchrotron radiation Is there an easy explanation why synchrotron radiation from a bending magnet (e.g. in an electron storage ring) has an energy distribution? In other words, given a specific magnet with a defined magnetic field and electrons with a specific kinetic energy, why is the produced synchrotron radiation not monochromatic?
| It’s a relativistic effect! For non-relativistic cyclotron radiation, the spectrum is mostly monochromatic at the gyro-frequency and the angular distribution is basically over all angles. But for highly relativistic synchrotron radiation, the radiation from the electrons is strongly “beamed”, like a headlight, in a very narrow cone ahead of the electron. This means that the power received by a distant observer is a sharp spike in time, once per revolution. The Fourier transform of this sharp spike contains a broad distribution of frequencies.
The graphs on page 52 of this PDF show the transition from the monochromatic non-relativistic case to the broad-spectrum highly-relativistic case. Note that these graphs have linear rather than logarithmic axes, so the final one doesn’t look like the one you showed.
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Angular velocity of rotating rod Consider the following system:
Newton's second law for rotational motion:
\begin{equation}\tau=I\alpha \Leftrightarrow rF=\frac{1}{3}mr^{2}\alpha \Leftrightarrow \frac{d\omega}{dt}=\frac{3F}{mr}\end{equation}
Considering RHS constant, we get $\omega=\frac{3F}{mr}t.$
I'm not sure if the angular velocity whould be inverse proportional to the radius (from natural experience I know that pushing farther requiers lower force).
Also what happens if the bar is not fixed and the two opposite forces are acting at the ends of the bar.
Since their sum is $\vec{0}$ there is translational equilibrum and so the axis of rotation is at the $C.M.$ but will the action of the two forces change the angular velocity from the previous situation?
|
I'm not sure if the angular velocity whould be inverse proportional to the radius (from natural experience I know that pushing farther requiers lower force).
You are forgetting that you are also using $r$ as the length of the rod. So if you increase $r$ you are increasing $\tau$ that scales linearly with $r$, but then you also increase $I$ which scales quadratically with $r$. So it makes sense that you end up with a $1/r$ for $\alpha$.
I know that pushing farther requiers lower force).
Also what happens if the bar is not fixed and the two opposite forces are acting at the ends of the bar.
Since their sum is $\vec 0$ there is translational equilibrum and so the axis of rotation is at the C.M. but will the action of the two forces change the angular velocity from the previous situation?
I would suggest just doing your previous analysis. Determine the net torque acting on the rod about the C.M., and then you can determine the angular acceleration about the C.M. using your rotational analog of Newton's second law.
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If an object is travelling near light speed, would it's actions seem to be in slow motion? Hypothetically if we were observing a clock travelling near light speed relative to us, we would see the clock ticking at a much slower speed than us. If that is true, then would all actions that are at rest relative to the clock seem to be slower too? For example if the clock were to explode, would we observe the explosion to be a slower speed?
If my question doesn't make sense then please ask for clarification I'm having trouble putting my thoughts into words.
| Due to time dilation , any system that would be moving with speeds near to the speed of light, would be slower. All actions would take longer time.
But if you were inside that system, ie also moving with speeds comparable to that of light, you will observe everything normal in your spaceship but the people on earth will appear to go slower for you
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Can a wire having a $610$-$670$ THz (frequency of blue light) AC frequency supply, generate blue light? We know that when we give alternating current across a wire then it will generate an electromagnetic wave which propagates outward.
But if we have a supply which can generate 610 to 670 terahertz of alternating current supply then does the wire generate blue light?
| Yeah definitely . You can create light corresponding on any frequency by this method. It’s just that creating this circuit will be very challenging. To give you a perspective, the highest frequency that we have obtained with modern electronic circuits is around 10^11 Hz.
| {
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Direction of emission of Photoelectrons Where does the information about the direction of the emission of the Photoelectron come from? Does it get it from the incoming Photon?
I have seen a picture on wikipedia-page of the photoelectric effect, where it almost looked like if the angle of the emission of the Photoelectrons were the same as the angles from the incoming Photons. Is there some "law" which describes why and how the Photoelectrons are emitted at a certain angle like there is for the reflection of light when it hits a reflecting surface in optics?
| The direction of emission of photoelectrons during the photoelectric effect is random. It is as per QM, all about probabilities.
When you are comparing it to a mirror image, that is not correct. A mirror image is caused by elastic scattering.
When a photon interacts with an atom, three things can happen:
*
*elastic scattering, the photon keeps its energy and changes angle
*inelastic scattering, the photon keeps part of its energy and changes angle
*absorption, the photon gives all its energy to the atom
A mirror image is built by 1., elastic scattering, that is the only way to keep the energy and phase of the photons.
The photoelectric effect is 3., absorption, when the energy of the photon is transferred to the atom and the electron gets kicked off, because the photon's energy level is enough to reach the work function of the electron.
The angle of the kicked off electrons is random.
| {
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Why electric potential requires absence of acceleration? As of 28/05/2019, Electric Potential is defined in Wikipedia in the article with the same name as:
The amount of work needed to move a unit of positive charge from a
reference point to a specific point inside the field without producing
an acceleration
What is the meaning of "without producing an acceleration"?
| The total energy of a body is the sum of kinetic energy and potential energy.
Kinetic energy depends on speed and potential energy depends on position.
The wiki text is a special case of the fact that if we move a body from A to B without altering the kinetic energy then its energy difference is only potential.
| {
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Unorthodox way of solving Einstein field equations Usually when we solve field equations, we start with a stress energy tensor and then solve for the Einstein tensor and then eventually the metric. What if we specify a desired geometry first? That is, write down a metric and then solve for the resulting stress energy tensor?
| In most situations the distinction “matter first, geometry second” or “geometry first, matter second” is not that clear cut. Often assumptions are made that constrain both the geometry and stress energy tensor.
Take for example the Schwarzschild metric. We derive it by writing down the most general metric compatible with restrictions imposed by physical description: “isolated static body with spherical symmetry”:
$$
ds^2=-A(r)dt^2 + B(r)dr^2 + r^2(d\theta + \sin^2 \theta d\phi^2).
$$
Only then we substitute this metric into vacuum Einstein equations (with zero stress energy tensor) and obtain a couple of ordinary differential equations for functions $A(r)$ and $B(r)$. So, we solve equations for a given matter content, but these equations are in a simple form because we specified large parts of geometry first.
Another class of examples are what could be called “science fiction geometries”: time machines, warp drives, traversable wormholes that challenge our intuition on what is allowed in the universe. Such “solutions” are often start from geometry written down with a desired properties but the Einstein field equations are still considered in order to constrain what form of “exotic matter” is needed to obtain such geometries. Parameters of the geometry are often varied in order to minimize the “unnaturalness” of the resulting stress energy tensor. A few examples:
*
*Alcubierre warp drive and its variations allows faster than light travel with the help of negative mass.
*Traversable wormholes would allow travel (or communication) between distant regions of the Universe (or between different universes). See this paper for an example of obtaining conditions of stress-energy for such a spacetime.
Yet another group of examples which have priority of the geometry over matter comes from astrophysics: observations often give us information about spacetime which could then be used to deduce the matter content. That is essentially how $\Lambda$CDM model appears, the matter content, most notably the dark energy is deduced from spacetime structure.
| {
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Is my understanding of vectors correct? I recently learned that a vector in mathematics (an element of vector space) is not necessarily a vector in physics. In physics, we also need that the components of the vector on a coordinate transformation as the components of the displacement vector change. So, if my understanding is correct, if $|\mathbf{c}_1|, |\mathbf{c}_2|, |\mathbf{c}_3|,\, \ldots \,,|\mathbf{c}_n|$ are the components of a vector $\mathbf{A}$ and $f$ is the function of transforming coordinates (change of basis), then $$f(\mathbf{A}) = \sum_{i=1}^n{f(\mathbf{c}_i)}$$ where $\mathbf{A} = \sum_{i=1}^n\mathbf{c}_i$.
That is to say, the transformed vector by applying $f$ to it should be equal to the vector formed by the vector components which have been transformed by applying $f$ to them.
Am I correct?
| Yes, your statement is correct, however... I haven't seen your use of "component" in a long time. Your use is the strictly correct meaning of component. That is, components are vectors. But the term is often used to mean the "coordinates" of a vector. That is in $\vec{v}=v_{x}\hat{i}+v_{y}\hat{j}+v_{z}\hat{k}$ the actual components are $v_{x}\hat{i}, v_{y}\hat{j}$ and $v_{z}\hat{k}$. But people almost always mean $v_{x}, v_{y}$ and $v_{z}$ when they say "components." And $f\left(v_{y}\hat{j}\right)$ (vector argument) and $f\left(v_{y}\right)$ (scalar argument) are not the same thing. In most situations a function (transformation) taking a vector argument will not be defined for a scalar argument.
| {
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Which one of them is the time-reversed wave-function, $\psi^{\ast }\left( x,t\right) $ or $\psi^{\ast}\left( x,-t\right) $? If the wave function $\psi\left( x,t\right) $ is a solution of the spinless
time-independent Schr$\ddot{\mathrm{o}}$dinger equation,
$$
i\hbar\frac{\partial}{\partial t}\psi\left( x,t\right) =\left[ -\frac
{\hbar^{2}}{2m}\nabla^{2}+V\left( \mathbf{r}\right) \right] \psi\left(
x,t\right)
$$
then, $\psi^{\ast}\left( x,-t\right) $ is also the solution
$$
i\hbar\frac{\partial}{\partial t}\psi^{\ast}\left( x,-t\right) =\left[
-\frac{\hbar^{2}}{2m}\nabla^{2}+V\left( \mathbf{r}\right) \right]
\psi^{\ast}\left( x,-t\right)
$$
and can be defined as the time reversed wave function of $\psi\left(
x,t\right) $
$$
\psi_{r}\left( x,t\right) =\psi^{\ast}\left( x,-t\right)
$$
However, in many discussions about the time-reversed operation, the time
reversed wave function $\psi_{r}\left( x,t\right) $ is obtained by applying
the time reversal operator $K$, which is the complex conjugate of the wave function,
$$
\psi_{r}\left( x,t\right) =K\psi\left( x,t\right) =\psi^{\ast}\left(
x,t\right)
$$
So my question is, which one is the time reversed wave function $\psi^{\ast
}\left( x,t\right) $ or $\psi^{\ast}\left( x,-t\right) ?$
The general expression for the time-reversal operator $T=UK$ (Eq. (4.4.14) in
Modern Quantum Mechanics by J. J. Sakurai), where $U$ is a unitary operator
and $K$ is the complex conjugation operator. For spinless case, one can choose
$U=1$, so $T=K$.
| As per your reference, it seems that you have mistaken anti-unitary operators for the time reversal operator. The time reversal operator is a kind of anti-unitary operator. The general expression for an anti-unitary operator is, as you had mentioned, on page 269 equation 4.4.14 of J.J Sakurai's book:
$$
\theta = U K
$$
Where $\theta $ is an anti-unitary operator, U is a unitary operator and K is the complex conjugation operator.
You can't simply take U as the identity, as even though this is an anti-unitary operator, it is not necessarily the time reversal operator.
| {
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Why does the Schrödinger equation work so well for the hydrogen atom despite the relativistic boundary at the nucleus? I have been taught that the boundary conditions are just as important as the differential equation itself when solving real, physical problems.
When the Schrödinger equation is applied to the idealized hydrogen atom it is separable and boundary conditions are applied to the radial component. I am worried about the $r=0$ boundary near the nucleus. Near the proton, the electron's kinetic energy will be relativistic and looking at the Schrödinger equation itself for how this boundary should behave seems dangerous because its kinetic energy term is only a non-relativistic approximation.
Is there any physical intuition, or any math, that I can look at that should make me comfortable with the boundary condition in this region?
| In solving the Schroedinger radial equation there is no boundary condition applied at $r=0$. At $r=\infty$ yes, $R(r)$ must tend to zero - so we reject the positive exponential solution; any change in that would have massive consequences. But there is no constraint laid on $R(r)$ or indeed $R'(r)$ as $r \to 0$.
So there's not a change in the boundary condition. There is a change in the kinetic and potential energies due to relativistic effects and the fact that the proton is not a point charge. These do have an effect - but very small, as the volume concerned is about $10^{-15}$ of the volume of the atom. (Actually atomic physicists experiments can detect these effects, at least for large $Z$ atoms, thanks to some very clever and precise optical experiments.) But this is a small effect, not the game changer that a new boundary condition could give.
| {
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Can Hydraulic System work on a Moon Robot? Since the Moon has no atmosphere and the temperatures reach maximum 123 C and min minus 153 C, how feasible is it to use hydraulic actuators to move the Robot legs?
Since my assistant Professor insists on going forward with the idea of building a smaller scaled model using the hydraulic system, i was a bit skeptic and wanted to know if it is actually doable.
For the record the Robot will run with batteries with max capacity of around 25000 mAh. I thought it was important to mention that because of the possibility to keep the hydraulics System within operation temperatures.
The robot should also be able to handle around 1000 N of load.
| yes hydraulics could be made to work, the lack of air pressure is no big problem since hydraulic pressures will be fairly high, a few missing psi externally wouldn't really matter. The biggest problem would be having a hydraulic fluid with a multi-viscosity to handle the extreme temperature variations. Also any hoses or seals would have to remain flexible and pressure proof at those extremes.
| {
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Can I contract index in this expression? I'm reading Carrol text on general relativity, on page 96 they arrive to the term
\begin{equation}
\frac{\partial x^{\mu}}{\partial x^{\mu '}}\frac{\partial x^{\lambda}}{\partial x^{\lambda '}}\frac{\partial^2 x^{\nu '}}{\partial x^{\mu}\partial x^{\lambda}}.\tag{1}
\end{equation}
Can I contract this expression to get
\begin{equation}
\frac{\partial^2 x^{\nu '}}{\partial x^{\mu '}\partial x^{\lambda '}}~?\tag{2}
\end{equation}
I'm using the chain rule $$\frac{\partial x^{\mu}}{\partial x^{\mu '}} \frac{\partial}{\partial x^{\mu}}=\frac{\partial}{\partial x^{\mu '}}\tag{3}$$ (which I think is correct).
| No, because $$\color{red}{\sum_{\mu,\lambda}}\frac{\partial x^{\color{red}{\mu}}}{\partial x^{\mu '}}\frac{\partial x^{\color{red}{\lambda}}}{\partial x^{\lambda '}}\frac{\partial^2 x^{\nu '}}{\partial x^{\color{red}{\mu}}\partial x^{\color{red}{\lambda}}}$$already has a double summation implied in $\mu$ and $\lambda$, so these are not free indices, and hence cannot be contracted with anything.
| {
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What is so special about Paschen, Balmer and Lyman series? I'm reading McIntyre's QM book and I see the same diagram ever so often:
He goes on to say:
I don't understand what he is trying to say with the last sentence. It seems he is saying, that transitions to higher-levels higher than n = 5 requires absorption of photons. But every level requires the absorption of photons. For example, to go from 1 to 2, we need light absorption.
*
*What does he mean here?
*Perhaps I am missing the significance of the three series. Perhaps there is some other physics going on?
*What is the significance of the three series? Why are they always shown in textbooks?
| 1 He is saying that, just as in the three series described above, the absorption of photons holds for any n.
2-3 The significance of these series is mainly historical, because they are of course related to the first energy levels. There is nothing particularly special about any of these series other than the experiments whence they come from. For instance, the Lyman series are important because they study the UV spectrum of the hydrogen gas, which by itself was a hot topic at the time.
You could benefit from reading about the history of these series, but just keep in mind that these transitions are no different from any others.
| {
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Monte Carlo Closure Test in Particle Physics What is the principle behind a MC Closure Test employed while analysing data in particle physics?
So, what I understand here is that this test checks if one's code is working properly or not depending on whether the (data-mimicking) MC, after being analysed, matches with the true MC information. Thus we have a closure and therefore the name: MC Closure Test.
Also, is using the same MC for analysing and then for final comparison a good idea?
Strange thing is that there is little publication that is easily available explaining the intricacies of the test.
This Q&A is probably related: https://physics.stackexchange.com/a/408495/46907.
| It is not a widely discussed subject , and the concept was certainly not used in my time ( twenty years ago) . I found this definition:
When using pseudo-data, generated with the help of Monte Carlo simulations, the truth distribution $x^{truth}$ is known, so the unfolding result $ˆx$ may be directly compared to it. Such comparisons, where pseudo-data are unfolded and compared to the truth are often called closure tests.
So I understand it is a comparison of theoretically known distributions, which are used to generate the "unfolding" result, and it should be within the estimated errors for the pseudo data .
This is a use in high energy data:
Closure Test for Cross Section Analysis
Closure means that the acceptance and efficiency corrections work
It is still too esoteric for me, lets hope that somebody who has used the method replies .
Also, is using the same MC for analysing and then for final comparison a good idea?
Monte Carlo is a method of integration. If one can do the integration analytically , one can compare with the simulated data which has statistical and systematic errors, and these should agree within errors. It is two different comparisons of monte carlo data. The objective is to catch and discrepancies within the integration, so that new physics would appear in comparison with data.
| {
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Why does steam in Chernobyl reactors speed up nuclear reaction? After watching the TV series Chernobyl I am struggling to understand why steam in a reactor core would increase the rate of the nuclear reaction. My first guess would be that liquid water would accelerate the reaction as it is used as a moderator in other reactors.
| The reactor type installed in Chernobyl (Wikipedia: RBMK) uses graphite as moderator. The water also moderates, but its effect is proportionally much less important than in a water-moderated reactor. This means a steam bubble forming results in only a small reduction in moderator efficiency but a large reduction in neutron absorption, thereby increasing the fission rate. This behaviour is referred to when the reactor is described as having a high positive void coefficient (the link goes to the subsection of the above Wikipedia article discussing this).
| {
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Tension in a string moving in elevator A lift can move in $x$ axis and also in $y$ axis. A bob of mass m is suspended with inextensible thread inside the ceiling of elevator. Determine tension $T$ in the string when elevator moves in $x$ direction.
Analysing the motion w.r.t. lift (considering that lift moves with acceleration $a$) the forces on the bob are tension upwards, $mg$ downwards and $ma$ left ( pseudo force). Equating forces in $y$ direction as there is no motion along $y$, I end up getting $T=mg$. Where have I done wrong?
The next question in same paragraph is what is the tension when lift moves down with constant velocity.
The only downward force on bob is $mg$ and upward force is $T$. Pseudo force is $0$ as the lift is moving with constant velocity. Again I am getting $T=mg$. Where I have gone wrong?
| Your free body diagram holds only for $t=0$.
Observe that at $t=0$, due to the force $ma$, there exists a torque on the bob, $\tau=ma\ell$ where $\ell$ is the length of the string. This causes the bob to rotate. If $a$ is directed to the left for the elevator, the inertial pseudo-force on the bob will be directed to the right, causing the bob to swing anticlockwise.
As long as $ma$ isn't too great, the bob will tend to reach an equilibrium, namely where the torque becomes zero. Let $\theta$ be measured from the vertical. Then the torque becomes zero precisely when:
$$mg\ell\sin{\theta}=ma\ell\cos{\theta}$$
Which can be shown with geometry. This reduces to:
$$\theta=\arctan{\frac{a}{g}}$$
Intuitively, observe that this makes sense. The larger the lateral acceleration, the larger we can expect the angle of deflection to be.
Since at this point, the bob is in equilibrium, the tension in the string must be equal to the magnitude of the vector sum of weight and lateral force. In other words:
$$T=m\sqrt{a^2+g^2}$$
We use the Pythagorean theorem since the lateral force and weight are always orthogonal.
For the second part, as far as I know, since the velocity is constant, $T=mg$. There are no torques at play here. The problem is entirely one-dimensional.
Please comment if anything seems off!
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How to prove a 4D vector is a 4-Vector? This is a fairly open ended question.
Given a set of 4 Components, that is, a 4D Vector, what is the process for determining rather or not it is a "4-Vector" as defined in special relativity? I want to know the general method for answering this question. It is still unclear to me what additional constraints make 4-vectors "special".
| A set of 4 components, that is a 4-dimensional vector, is a 4-vector if it is transformed as the position 4-vector between reference frames.
Related : Transformation of 4−velocity.
| {
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If the pressure inside and outside a balloon balance, then why does air leave when it pops? Sorry for the primitive question but when we inflate a rubber balloon and tie the end, its volume increases until its inner pressure equals atmospheric pressure.
But after that equality is obtained why does the air goes out when we pop the balloon? If there is pressure equality what causes the air flow?
| The pressure inside is greater until it balances with external pressure. The rubber is the only thing maintaining the balance. If it were any greater inside, it would continue to expand.
It takes very little pressure to lift large weight when applied to large surface areas. Since this minor pressure it applied to the entire inner surface it takes very little pressure to maintain.
Inflation takes pressure, pressure has to balance or inflation continues. It's not a steel belted tire. How much pressure does a thin rubber membrane take? How small a flaw destroys it?
I actually untied a helium balloon and let it go many years ago and it did not deflate. AS IT floated up we all could see into the balloon through the opening as it lifted, but did not deflate at all. It continued til out of sight.
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Would setting the ideal gas constant to $1$ yield an attractive natural temperature scale? In this recent question, there was a comment 'The "zero point" of Kelvin is natural, but the scale is not'. This led me to wonder whether setting $R = 1$ in the ideal gas law would be an attractive and more natural temperature scale.
I am aware that changing to such a scale is not practical, the investment in the Kelvin is too great.
|
Would setting the ideal gas constant to 1 yield an attractive natural temperature scale?
Not really. The universal gas constant involves two arbitrary units: energy and temperature, and also the unitless mole. Getting rid of the concept of moles results in Boltzmann's constant $k_b$. This is the value that is set to 1 rather than $R$ in all systems of natural units.
| {
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If a black colored body absorbs all colors, why does the spectrum of light appear on a black shoe? Few days back in a 10 grade school practical, we were shown the dispersion of light by a prism to create spectrum.
Then we went into the open sun and performed it under a linear building roof and same results were obtained.
But when a teacher took it in the light and the spectrum was displayed on the ground cement, a child's shoe can in the spectrum's way and the spectrum was still visible on his black shoe.
Now my question is if the objects that appear black absorb the colors and radiate them back.
So why does the spectrum appears on the shoe when the colors should be absorbed by the shoe material and no color but black must appear as it is the absence of any color radiated?
And also i want to ask that do blackbodies radiate back the same color the absorb?
| Black object "absorb the light falling on them". Sure. But how much of the light is actually absorbed?
Your visual system works mostly on difference between stimulii, so "black" is usually incontrast to things that are less black. This is the origin of the Chubb illusion and the Checker shadow illusion.
Even very black day-to-day materials reflect some light. For instance copy tonor reflects a couple of percent of light.
Your classmates "black" shoe absorbs enough light that it is darker than everything you look at most of the time, but it does not absorb all the light, and in a dim space your eye is sensitive enough to see the small amount of reflected light.
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Tearing paper by creasing/folding Why does folding/creasing loosen the fibre-fibre bonding in paper?
Creasing makes tearing paper easier because it weakens the fibre-fibre bonds or makes the strong fibres easier to tear, it is said. But why would the creasing make the bond easier to break?
Some sources say that a certain elongation occurs at the sight while others mention about a deformation. Please explain.
| Paper consists of cellulose fibers held together with a small amount of binder (a glue-like substance). Creasing a piece of paper back and forth applies nonuniform shear and tensile forces to the fibers which tends to break the mechanical bonds between adjacent cellulose fibers. This then allows them to tear free of one another when you then unfold the paper and pull on it.
The strongest papers are those made from the longest fibers because they are more easily entangled with other fibers during the manufacturing process, and have the most binding agents. The weakest papers are those consisting of short fibers (and a minimum of binders).
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How do phone loudspeakers work? From what i understand, loudspeakers require AC signals to cause an electromagnet to oscillate due to changes in current direction, thus force direction. How can this happen with a phone's battery/cell? shouldn't the battery only be capable of producing direct current?
| Starting out from a speaker assembly, we need to generate a voltage that varies at the same rate as we want our membrane to vibrate and thus create sound. This problem also exists in AC powered devices, as the AC comes in at strictly a single frequency, which could generate only a single frequency of sound.
To do this, all digital devices use digital-to-analog (DAC) converters, almost always on a chip. They take some digital input signal at some of their pins, and output a voltage proportional to the value of that digital input on another pin. If we do this fast enough, i.e. feed different numbers in rapid succession, we will get a time-varying output signal.
If we happen to feed it input that corresponds to some audio signal we want to produce, and couple the output pin to a speaker, we are generating sound.
Edit: This is technically not yet alternating current, but AC on top of a DC signal. To get a real alternating current, the output of the converter can be connected through a capacitor (acting as a high-pass filter), which removes the DC part, so only AC can pass through.
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Stokes's theorem in tensor field On pg 73 of "Tensors, Relativity and Cosmology"
The generalized Stokes's theorem in arbitrary $N$-dimensional space is given by:
$$\int_c A_mdx^m=\frac{1}{2}\int_S F_{mn}dS^{mn} \tag{1}$$
where $F_{mn}$ is the curl tensor of the vector $A_m$, $F_{mn}=A_{n,m}-A_{m,n}$ (, denotes covariant differentiation here) and $dS^{mn}$ is the contravariant tensor of an infinteseimal element of the surface $S$ $(dS^{mn}=dx^m \wedge dx^n$).
In the three-dimensional metric space the RHS of (1) is equivalent to the ordinary curl A definition
I tried to expand the RHS of (1) to obtain
$$\frac{1}{2} \left(\frac{\partial A_n}{\partial x^m}-\frac{\partial A_m}{\partial x^n} \right)dx^m \wedge dx^n$$
since the Christoffel symbols vanish in the three-dimensional Euclidean metric space.
It appears that
$$\frac{\partial A_n}{\partial x^m} - \frac{\partial A_m}{\partial x^n}$$
is the definition of curl A but how do I convert $\frac{1}{2}dx^m \wedge dx^n$ into dS to obtain Stokes's theorem in the ordinary vector notation?
| Here, you still have the curl in antisymmetric tensor form, not a vector form. Once you are in three dimensions and there are no peculiarities with the metric, you have the correspondence:
$$(\operatorname{curl}{\bf A})_i=\frac12\epsilon_{ijk} \left(\frac{\partial A_{j}}{\partial x^k}-\frac{\partial A_{k}}{\partial x^j}\right) $$
where $\epsilon_{ijk}$ is the Levi-Civita tensor. The inverse of this expression is:
$$\epsilon_{inm}(\operatorname{curl}{\bf A})_i= \left(\frac{\partial A_{n}}{\partial x^m}-\frac{\partial A_{m}}{\partial x^n}\right) $$
When you use this on your expression, you get
$$\frac12\epsilon_{inm} dx^m \wedge dx^n$$
which is just the cross product you require to get to $dS_i$. Note that antisymmetric property of $\wedge$ goes together nicely with antisymmetric Levi-Civita and the one-half in front.
In terms of differential geometry, you convert a 2-form into 1-form with the Hodge star operator, which is a generalized and more formally correct version of what we did above.
| {
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Which particle mediates the Aharonov-Bohm effect? BACKGROUND
The Aharonov-Bohm (AB) effect induces phase shifts between the two paths that an electron could take around an enclosed magnetic field. In radial coordinates, assume that the magnetic field is localized around the origin and that the two paths traced by the electron form two complementary half-circles at radius some R. Assume further that the magnetic field is initially switched off.
QUESTION
At the moment the magnetic field is switched on, which particle travels outward from the origin towards the electrons' path so as to mediate the phase shift? And at what speed? Clearly, such a particle can't be a disturbance of the electromagnetic field since the magnetic field is restricted to the origin and its vicinity.
|
At the moment the magnetic field is switched on, which particle
travels outward from the origin towards the electrons
You have already answered yourself: magnetic fields are mediated by photons.
And at what speed?
Photons generally travel at the speed of light :-)
the magnetic field is restricted to the origin and its vicinity
Whoa, this is the issue right here. The range of any magnetic field is infinity. Its strength falls off, but there is no place where it is zero.
I am curious why you believe otherwise, as it seems you are aware that EM is due to photon exchange, and as photons have no half-life, one would naturally assume there is no inherent range limit (unlike, say, the strong force where the mediators have a short life and therefore can't get very far).
| {
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Swinging Atwood and Hoop And Pulley Lagrangian
The picture is showing the swinging atwood and a hoop and pulley.
I know the lagrangian for both two, I have no problem with the kinetic energy of both but i couldn't convince myself that for the swinging atwood, the potential energy is :
$$V = Mgr - mgr \cos\theta$$
and for the hoop and pulley, the potential energy is :
$$V = -Mgr \cos\theta - mgr\theta$$
where $\theta$ is the angle mass $m$ makes with vertical
I'm having the sign problem here. I don't understand why for the atwood we use $+Mgr$ and the hoop we use $-mgr\theta$
When and how do we use the $+$ and $-$ sign?
| In the first case we have $Mgr$ because as $r$ increases the potential energy increases for the block of mass $M$. In the second case we have $-mgR\theta$ because when $\theta$ increases the potential energy decreases for the block of mass $m$.
In general you pick the sign so that the potential energy function behaves how it should given the variables it depends on.
| {
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Shooting someone's past self using special relativity Suppose A and B are a long distance apart initially. B takes off in a spacecraft in A's direction at a really high speed. Both were aged 0 when B took off. When B about to cross paths with A, A observes him to be 30 years old (while A is 60). At this point, 60 year old A shoots the 30 year old B. B dies.
Now, from B's frame, the bullet has to be fired by A when B is 120. That's because the event of 'bullet firing' must happen after the event of 'A turning 60', and A turns 60 in B's frame only when B is 120. So B has to die at 120 as seen from his own frame of reference.
How can B die at both 30 years old and 120 years old?
| Case I: When B takes off, A and B are both aged 0 in A's frame.
1) When B passes A, B is 30 (given in your setup).
2) But B's clocks run at half-speed according to A, so A says that B has been traveling 60 years.
3) Therefore A is 60.
4) But B says A's clocks run at half speed, so B says A was born 120 years before the shooting --- that is, 90 years before B was born.
5) So B's story is this: "90 years before I was born, A was born. He aged at halftime, so on my birthday, he was 45. At that time I started my journey to earth, which took 30 years. During that time, A aged another 15 years, so he was 60 when we met. Then he shot me. I died at 30."
Your mistake: You said that "from B's frame, the bullet has to be fired by A when B is 120". That's not correct. The correct statement is "from B's frame, the bullet has to be fired 120 years after A was born". Since B is 30 at the time of the shooting, A must have been born 90 years before B.
Your bigger mistake: You assumed that two different problems (namely this one and the one you asked in your last post) have to have exactly the same answer. In the other problem, A and B were in the same place at the same time when both were born. In this problem they weren't.
Case II: When B takes off, A and B are both aged 0 in B's frame.
1) When B reaches earth, he is aged 30 (given in the problem).
2) According to B, A ages at half-speed. Therefore A is 15, not 60 as you supposed. The 15-year-old A shoots the 30 year old B. Game over for B.
| {
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How does a Leyden Jar work? When electricity goes inside the Leyden jar, the inside of the jar is negatively charged. Then the aluminum foil on the outer layer of the jar is positively charged. Does the inside of the jar want to cancel out the aluminum foil, but it can't because of the insulator between the two conductors?
| All you have in a Leyden jar are two thin tin foils on the interior and exterior of the jar.When you pass a current like you said one plate is going to have -q amount of charge and the other will have +q amount of charge.But they cant cancel out each other since the have a insulator in between that doesn't allow the charges flow.
| {
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Is there a maximum distance from a planet that a moon can orbit? Given a planet that orbits a star, and a moon that orbits that planet, is it possible to define a maximum orbital radius of that moon, beyond which the moon would no longer orbit the planet, but the star instead?
I initially (naively) thought this point would be where the star's gravity outweighed that of the planet:
$$d_\text{max} = d_\mathrm p - d_\mathrm px$$
$$x = \frac{1}{\sqrt{\frac{m_\mathrm p}{m_\mathrm s}}+1}$$
Where:
$d_\text{max} = $ maximum orbital radius of the moon (around the planet), $d_\mathrm p =$ orbital radius of the planet (around the sun), $m_\mathrm p =$ mass of the planet, $m_\mathrm s = $ mass of the star.
But I quickly realised this assumption was wrong (unless my shoddy maths is wrong, which is very possible), because this gives a value of $258\,772\ \mathrm{km}$ using values of the Sun, Moon, and Earth. $125\,627\ \mathrm{km}$ closer to the Earth than the Moon's actual orbital radius (values from Wikipedia).
Is there a maximum orbital distance? How can it be calculated?
| What you want to look for is the Hill Sphere. The resultant distance is about 1.5 million km.
| {
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Is it possible writing conservation of relativistic energy in this naive way? Conservation of charge or rest mass can be written in this way and it is Lorentz invariant
$$
\nabla \cdot (\rho \mathbf{u}) + \frac{\partial \rho}{\partial t} = 0
$$
So we could be tempted to naively write conservation of energy in this way (I use $\gamma_u$ for particle in motion at speed $\mathbf{u}$ to not making confusion with $\gamma$ relative to speed of $S'$)
$$
\nabla \cdot (\gamma_u \rho \mathbf{u}) + \frac{\partial (\gamma_u \rho)}{\partial t} = 0
$$
But this doesn't look Lorentz invariant. I wrong? Exploiting vector identity $\nabla \cdot (\Psi \mathbf{A}) = \Psi (\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla \Psi)$ (whit $\Psi=\gamma_u $) and exploiting conservation of mass, this equation became
$$
\left(
\mathbf{u} \cdot \nabla + \frac{\partial}{\partial t} \right) \gamma_u = 0
$$
where mass is strangely disappeared. But transforming the corresponding primed equation with
$$
\frac{\partial}{\partial x'} = \gamma \left( \frac{\partial}{\partial x } + \frac{v}{c^2} \frac{\partial}{\partial t } \right)
$$
$$
\frac{\partial}{\partial y'} = \frac{\partial}{\partial y}
$$
$$
\frac{\partial}{\partial z'} = \frac{\partial}{\partial z}
$$
$$
\frac{\partial}{\partial t'} = \gamma \left( \frac{\partial}{\partial t } + v \frac{\partial}{\partial x } \right)
$$
$$
u_x' = \frac{u_x - v}{1-\frac{u_x v}{c^2}}
$$
$$
u_y' = \frac{u_y}{\gamma \left( 1-\frac{u_x v}{c^2} \right)}
$$
$$
u_z' = \frac{u_z}{\gamma \left( 1-\frac{u_x v}{c^2} \right)}
$$
$$
\gamma_{u'} = \gamma \gamma_u \left( 1 - \frac{u_x v}{c^2} \right)
$$
we get
$$
\left(
\mathbf{u} \cdot \nabla + \frac{\partial}{\partial t} \right) \left[ \gamma_u \left(1-\frac{u_x v}{c^2} \right) \right] = 0
$$
That it is different than $\left(
\mathbf{u} \cdot \nabla + \frac{\partial}{\partial t} \right) \gamma_u = 0$ written above. Another road could be exploiting
$$
\frac{\partial \gamma_u}{\partial x_i} =
\frac{\gamma_u^3}{c^2} \mathbf{u} \cdot \frac{\partial \mathbf{u}}{\partial x_i} \qquad \textrm{where $x_i=x,y,z,t$}
$$
to transform $\left(
\mathbf{u} \cdot \nabla + \frac{\partial}{\partial t} \right) \gamma_u = 0$ into
$$
\mathbf{u} \cdot \left(
\mathbf{u} \cdot \nabla + \frac{\partial}{\partial t} \right) \mathbf{u} = 0
$$
but this equation too doesn't lead to the invariance (although $\left(
\mathbf{u} \cdot \nabla + \frac{\partial}{\partial t} \right) \mathbf{u} = 0$ is actually invariant). There is a way to check the invariance, or writing conservation of energy in that simple way is incorrect?
| We can't just insert $\gamma$ into the equation of continuity, which in this case is a formulation of conservation of rest mass for a free particle:
$$
\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf v) = 0,
$$
and expect the resulting equation will still be valid.
Also, although the above equation has the same form in all inertial frames, this by itself does not imply that the 4-tuple $(\rho c, \rho\mathbf v)$ is a four-vector. In this case it is a 4-vector, similarly to electric current density $j^\mu$. But there are other cases where the same kind of equation $\partial_\mu S^\mu = 0$ is valid in all frames, but where $S$ is not a four-vector. Notable example is the Poynting energy density and momentum density 3-vector in matter-free space.
Similar things will happen for matter energy; even if (and that is a big if) one could derive such simple equation for this energy, this wouldn't imply the energy 4-tuple is a 4-vector. In fact, in EM theory based on Maxwell's equations there is no way to formulate energy conservation where energy density of matter or EM field is a part of some 4-vector field; one must one 4-tensors of 2nd rank (which are represented by 4x4 entries).
| {
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Why is the drag force proportional to $v^2$ and defined with a factor of $1/2$? $$Drag = \frac{1}{2}C_d \rho Av^2$$
I understand that the strength of the drag depends on the density of the fluid the body passes through, the reference area of the body, the drag coefficient, and the velocity of the object.
I don't, however, understand the 1/2 and the $v^2$ in the equation.
| It's similar question if you would ask - Why kinetic energy is defined as $E_k = 1/2~m v^2$. The answer is below.
Elementary work is :
$$
\begin{align}
dW &= F \cdot dr
\\&= F\frac {dr}{dt} dt
\\&=Fv~dt
\\&=m\frac {dv}{dt}v~dt
\\&=mv~dv
\end{align}
$$
Now integrate both sides :
$$ \int dW = m \int v~dv $$
Which gives :
$$W=E_k=1/2~mv^2$$
Thus the answer of why kinetic energy is proportional to $v^2$ and has half factor 1/2 in it, is that it is due to integration.
| {
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Partition function of Hamiltonian without momentum dependence Considering an ideal gas of $N$ diatomic molecules with dipole momenta $\vec{p}$. Given the Hamiltonian of one molecule
$$ H = -\vec{p} \cdot \vec{E} = -pE\cos(\theta)$$
when calculating the partition function we "ignore" the p-dependence by not integrating over the momentum space, but is this really right? How is this justified? Do we just "ignore" every variable, which the Hamiltonian is not dependent on? This is what I would have done:
\begin{align}
Z & = \frac{1}{h^3} \int \mathrm{d}^3q\ e^{\beta p E \cos(\theta)} \\
& = \frac{1}{h^3} \int_0^R \mathrm{d}r \int_{-\pi}^\pi \mathrm{d}\varphi \int_0^\pi \mathrm{d}\theta\ r^2 \sin(\theta) e^{\beta p E \cos(\theta)} \\
& = \frac{4\pi R^3}{e\beta h^3 p E} \sinh(\beta p E)
\end{align}
When using spherical coordinates to compute the partition function of the given Hamiltonian, do we (as a consequence) not integrate over the radius $R$ and $\varphi$?
EDIT: added my calculation
| Yes, we cannot ignore the momentum dependence of the system and must integrate over all of state space. However as @d_b pointed out, this is not the momentum, it is the dipole moment. So, this is not the same $\theta$ associated with the position of the centre of mass of the diatomic molecule but rather the angle between the dipole moment and the electric field.
In the partition function, we have to sum over all possible states of the system. In the above Hamiltonian, there is no dependence on momentum. Perhaps, this is a special case of a sort of lattice. We can add a momentum dependence to the Hamiltonian(a $\frac{\vec{p}^2}{2m}$ term.
Since there is no coupling between the momentum and the dipole moment, the partition function for a diatomic molecule will separate into 3 parts:
$$
z = z_{dipole} \times z_{momentum} \times z_{position}
$$
First we can find the dipole part that is:
$$
z_{dipole} = \int_0^\pi Exp(-\beta \; H(\theta))\;g(\theta) \; d\theta
$$
Here $g(\theta)$ is the degeneracy of the diatomic molecule. If we assume that this degeneracy is constant g then we can simplify the partition function by evaluating the integral.
For the other 2 partition functions, we can use the results of the ideal gas.
I have not evaluated any integral but i hope i have answered your question! Please let me know if you would like to know more or have a query!
| {
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What makes north pole of a magnet north pole in the first place? This question might seem absurd and illogical to many. But it just popped out in my mind while I was reading about magnetism.
-Like in case of charges, positive and negative charge on an atom means absence and presence of extra electrons respectively. So my question is what aspect exactly makes a pole of magnet north or south?
Is it absence or presence of something?
-I asked my teacher about this and he simply replied that north pole is something which attracts south pole. But this is more of a property to me rather than an exact meaning of what exactly is north pole of a magnet.
| To answer that you need to know that in a classical mechanic view elementary quantities of magnetic field are generated by small loops of current call magnetic dipoles. Macroscopic magnet are only an assembly of elementary magnetic dipoles in macroscopic dipole. Although the magnetic field generated by permanent magnet is purely quantum physics (so moving charges are not define) the analogy remain acceptable. The direction of the magnetic field is only define by the direction of current in the loop (clockwise or anti-clockwise), by reversing the current you invert poles. So north pole is define as the pole above a loop of "clockwise" current (an below anti-clockwise). The field goes out the north pole of a magnet and goes in the south pole of the magnet. This property come from link between current and magnetic field : Biot-savart law (or Ampere-Maxwell equation) which describe the magnetic field as the vector product of the current and the vector joining the elementary current element and the point as which the field is being compute.
Like for the electrostatic where positive (or negative) charged macroscopic object mean absence (or present) of electron, a macroscopic magnetic object indicate present of extra dipoles from on kind (clockwise or anti-clokwise). north and south pole will be define by those in excess.
edit: "of course" contrary to the charge where it exist two kind of charges (positive and negative) the two kind of dipoles are the same but flip upside down. The clockwise or anti-clockwise type of a dipole is not an intrinsic property but can be "easily" change. A "strong" magnetic field can flip the internal elementary dipole of a permanent magnet and consequently reverse the pole of the magnet.
| {
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Magnetic field of an infinite hollow cylinder (with volume current) Consider an infinite hollow cylinder with inner radius $a$ and outer radius $b$. The volume current density flows anti-clockwise across the surface of the cylinder ($\vec{J} = J\hat{\phi}$). The charge density is $0$ everywhere.
How can I compute the magnetic field $\vec{B}$ for all points in space? It's fairly simple when the current flows parallel with the axis of the cylinder, for you can then use Ampère's law with an Amperian loop that's perpendicular to the current. In this case, I can't seem to think of a good loop. What I do know is that the field inside the cylinder is in the positive z-direction and outside in the negative z-direction.
| Compare the scenario with that of an infinite solenoid.
(Image source, note that $B \neq \mu nI$ in this case.)
Since you know the volume current density, you can calculate the current enclosed by the loop.
Also, the magnetic field outside (e.g. at point P) the infinite coil is zero, due to the following reason.
(image source)
For two loops far away, the resultant magnetic field at the point P in the figure is directed upward, and for nearby loops it directs downward. The net magnetic field becomes zero (This can be shown rigourously by integrating components of magnetic field from each loop).
So you can find the magnetic field inside the coil.
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Problem while constructing langmuir probe I am working on DC glow discharges and want to construct a langmuir probe. Circuit I am using is as shown in the picture
Also, I am applying ~1KV across cathode and anode.
Problem I am facing is that I am getting discharge btw probe and cathode. Which isn't unexpected, but destroys the purpose of probe, but then how to construct langmuir probe for DC glow discharges? What should be the appropriate circuit?
| you probably need a current limiting resistor in the part of the circuit containing the probe, either that or a way of isolating the probe ground from the power supply ground.
| {
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What is the change of entropy for a resistor at constant temperature? A 10 Ω resistor is held at a temperature of 300 K. A current of 5 A is passed through the resistor for 2 minutes. Ignoring changes in the source of the current, what is the change of entropy in (a) the resistor and (b) the Universe?
My attempt:
$\Delta Q=I^2 R t=5^2\times 10\times 2\times 60=30000\ J$
$\displaystyle\Delta S=\frac{\Delta Q}{T}=\frac{30000}{300}=100\ JK^{-1}$
Won't $\Delta S_\text{univ}$ assume the same value of $100\ JK^{-1}$?
| In this problem, it would be incorrect to say that the change in entropy of the resistor between its initial and final states is anything other than zero. Entropy is a function of state, and the initial state of the resistor (300K) is exactly the same as its final state (300K).
The fallacy in applying the expression $\int{dq/T}$ to determine the entropy change of the resistor in this problem would be that the process the resistor experiences is irreversible, and this expression can only be used to determine the entropy change for a reversible processes. For an irreversible process, one must first devise an alternate reversible path between the same initial and final states, and then determine the value of the integral for that path. If we follow this procedure for a solid, like our resistor, we find that $$\Delta S=MC\ln{(T_2/T_1)}$$where M is the mass of the solid, C is its heat capacity, $T_1$ is the temperature in the initial state, and $T_2$ is the temperature in the final state. And if, as in our problem, $T_2=T_1=300K$, $\Delta S =0$.
Here is the mechanistic explanation of what takes place: In the present irreversible process, our system, the resistor, receives work W from its surroundings (electrial work) and it dissipates this work irreversibly, returning an equal amount of heat Q to its surroundings. This dissipation of electrical work within the resistor translates into generation of entropy. But, the generated entropy does not stay in the resistor. If it transferred via the heat flow Q to the surroundings. So the net effect is no entropy change for the resistor.
With the entropy change of the resistor being zero and the entropy change for the surroundings being 100 J/kg,the entropy change for the universe is $$\Delta S_{universe}=\Delta S_{syst}+\Delta S_{surr}=0+100=100\ J/K$$
As expected, for this irreversible process, the change in entropy of the universe is positive.
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Understanding the rolling constraint for one cylinder rolling inside another cylinder This is the problem to find the equation of motion of 2 cylinders in which 1 cylinder is placed inside another cylinder with larger radius as shown in figure. The condition is that both are rolling. Now, in the figure they say:
The rolling constraint on 2nd cylinder can be written as :
$ r_{2} \phi_{2} = r_{1} (\phi_{1}+\theta} $
(Eqn 2 at last in picture)
With symbols as shown in figure below. I want to understand how they come up with this rolling condition(i.e formula).
Thanks in advance
| The $r_2 \phi_2$ is simply the length of the red line on the inner cylinder in your picture. That must be equal to the length of the red line on the outer cylinder, since this is the line that the inner cylinder tracked on the inner surface of the outer cylinder. That length is given according to the picture by $r_1(\phi_1+\theta)$ since the arc defining the red line is given by $\phi_1+\theta$ as can be seen from the picture and $r_1$ is the radius of the inner circle of the outer cylinder.
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Friction due to pure rolling on an inclined plane When a body is executing pure rolling we know that the point of contact of the body with the ground is at rest with respect to the ground. If that's the case no friction should act as it is stationary.So when a body is rolling down an inclined plane its point of contact is stationary , then how does friction act to cause a torque, as static friction only acts when there is a tendency of relative motion with respect to the ground.
| False premise. Static friction only acts when there is no motion between the two objects being considered in the system. For instance, a stationary block on an inclined plane is held there by static friction which equals the component of its weight down the plane, $mg\sin{\theta}$. Under such scenarios, the force of static friction is bounded by:
$$F_f\leq\mu_sN$$
Where $\mu_s$ is the coefficient of static friction and $N$ is the normal force.
You might be thinking of kinetic friction, which can only occur when the two objects in contact have relative motion.
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Anomalous dimension of double-trace operators Is it true that if a single-trace operator, say, $O$ acquires an anomalous dimension $\gamma_o$, then the anomalous dimension of the double-trace operator $O^2$ is $2\gamma_o$? If no, can anyone please provide counter-examples?
| A correct statement is that in the large-N limit of a theory, if you have an operator $O$ with scaling dimension $\Delta_O$, the theory will also contain an operator $O^2$ with scaling dimension $2\Delta_O$.
But be aware that:
1) This is a statement about scaling dimensions, not anomalous dimensions.
2) If you include 1/N corrections the relation $\Delta_{O^2} = 2 \Delta_O$ won't be true anymore.
| {
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"url": "https://physics.stackexchange.com/questions/488331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does friction always oppose motion? Recently I had the following misconceptions:
*
*Static friction always opposes the motion of body.
*The force of friction cannot initiate motion in a body.
Now I came to know that my understanding was wrong and that friction indeed can cause motion in bodies, and that static friction does not always oppose the motion of a body.
But I found this quite bizarre, how can a force which has always been taught to us to oppose motion, oppose points 1. and 2.?
| The top answer given is correct, but I wanted to slightly extend this:
how can a force which has always been taught to us to oppose motion, oppose points 1) and 2)?
Imagine a table, I slide a very heavy hockey puck over the table. Without friction, the puck keeps sliding. But because there is friction, the puck is slowed down.
Now let's put the table on wheels and do the same test again. Without friction, the table does not move and the puck slides right off. BUT with friction, the friction takes the kinetic energy of the puck and imparts it to the table. Because of this (and because we put the table on wheels), the table now starts moving forward.
Because there is friction, the kinetic energy is transferred from the puck to the table, which can cause the table to start moving (given the right circumstances whereby the energy imparted to the table overcomes the table's own friction with whatever it's resting on).
There are common real world examples here, e.g. someone who takes a running jump, and lands on a stationary carpet/skateboard, which then starts moving because the friction keeps the carpet/skateboard and the person's feet together and therefore the person's kinetic energy is (partially) transferred to the carpet/skateboard.
| {
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Why is it that when a chalk board gets cleaned, the area that used to have chalk is the cleanest? Why is it that when you erase a chalk board, the area where the chalk used to be becomes the cleanest? By that I mean that when you erase a chalk drawing, the board gets smeared with chalk dust, but the area where the drawing used to be has less dust on it than the rest of the board.
For example: In the first picture below I draw a simple chalk smiley face. Here the face is noticeable because it is the area with the most chalk. For the second picture, I erase it. You can still make out the picture, but notice that you recognize it because it is now the area with the least chalk.
I would expect that if chalk was stuck to a certain region of a chalk board, then after erasing it, some chalk residue would remain, but instead it seems like the opposite happens. I don't have a good answer for this problem.
| My guess.
The abrasive character of the chalk itself results in a higher coefficient of kinetic friction between the chalk and the board, than between the soft material eraser and the board, whereas the normal force applied to the board is the same in each case. The greater friction force takes away more material from the board surface.
Would you expect the board to be cleaner with or without sandpaper on the surface of the eraser?
Hope this helps.
| {
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Where the derivative corrections come from in Wilson renormalization? I known that in the Wilson renormalization process fast modes are integrated out in order to define an effective action for the low modes field. Considering phi to the fourth theory it's easy to see how the quadratic and quartic terms are corrected at 1-loop. These corrections lead to couplings renormalization but I don't understand what term in the fast modes path integral leads to the field strength renormalization Z. In other words where are the derivative interactions?
More precisely. If I define $\bar{\phi_0} =\phi_0 + \hat{\phi}_0$, in which $\hat{\phi}_0$ is the fast modes field, the path integral over $\hat{\phi}_0$ is:
$$\int D\hat{\phi}_0 \, e^{-\int d^Dx [\frac{1}{2}(\partial\hat{\phi}_0)^2 +\frac{1}{2} m_0^2 \hat{\phi}_0^2+\frac{\lambda_0}{4!}( \hat{\phi}_0^4 + 4\phi_0^3\hat{\phi}_0 + 4 \phi_0\hat{\phi}_0^3 + 6 \phi_0^2\hat{\phi}_0^2)]}$$
Considering the free action as $S_0=\int d^Dx \, \frac{1}{2}(\partial\hat{\phi}_0)^2$ I can expand the exponential in this way:
$$\int D\hat{\phi}_0 \,e^{-S_0}(1-\frac{1}{2} m_0^2 \hat{\phi}_0^2 -\frac{\lambda_0}{4!}( \hat{\phi}_0^4 + 4\phi_0^3\hat{\phi}_0 + 4 \phi_0\hat{\phi}_0^3 + 6 \phi_0^2\hat{\phi}_0^2) + \dots) $$
Now for example from $\phi_0^2\hat{\phi}_0^2$ I obtain the first correction to the 2-point function and so on. In order to calculate $1+\Delta Z$ and write the effective action as:
$$S_{eff}[\phi_0]=\int d^Dx\,[\frac{1}{2}(1+\Delta Z)(\partial \phi_0)^2 +\frac{1}{2} (m_0^2+\Delta m^2)\phi_0^2+\frac{1}{4!}(\lambda_0 + \Delta \lambda)\phi_0^4 + ...]$$
I need a term in the expansion which contains $(\partial\phi_0)^2$, right? Where is it?
| Derivative corrections appear in just the same way as usual. For example, quadratic terms with derivatives can be written as $\phi f(\partial) \phi$ where $f(\partial)$ is some combination of partial derivatives, such as $\partial^2 + m^2$ in free field theory. The corresponding propagator is $1/f(ip)$.
So derivative terms just correspond to nontrivial momentum-dependence of the propagator. You can directly compute the propagator in the Wilsonian picture (i.e. taking an expectation over fast modes). Any nontrivial dependence on the external momentum whatsoever corresponds to derivative terms.
In $\phi^4$ theory there are no quadratic derivative terms created at one-loop, because the one diagram you can draw doesn't actually have the external momentum flowing through the loop at all. But derivative terms do appear at two loops; they're totally generic.
| {
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How do I transform flux density into temperature? I have flux density data from Planck mission (in Jy) - measure in a certain frequency - and I would like to transform it to temperature data (Kelvin). I'm not sure if I should use Stefan Boltzmann's Law or not because the units don't match.
| We start from the Planck's law, describing the spectral density of electromagnetic radiation:
\begin{equation}
B_{\nu}(T) = \dfrac{2 h \nu^3}{c^2} \dfrac{1}{\mathrm{exp} \left( \dfrac{h \nu}{k T} \right) -1}
\end{equation}
Where $k$ is the Boltzmann constant, $\nu$ the frequency of the signal and $c$ the speed of light.
A Taylor expansion of Planck's law with $h \nu \ll kT$ (which is a valid assumption in radioastronomy) yields the Rayleigh-Jeans approximation:
\begin{equation}
B_{RJ}(\nu, T) = \dfrac{2 \nu^2}{c^2} k T
\end{equation}
With this relation a given intensity $I_{\nu}$ [Jy] can be linked to a blackbody temperature $T$ [K]. Indeed, this is the temperature of a blackbody that would emit an amount of light given by $I_{\nu} = B_{\nu}(T)$:
\begin{equation}
T = \dfrac{\lambda ^ 2}{2 k} I_{\nu}
\end{equation}
Also, it is possible to convert the integrated quantity $\int T dv$ to $\int I_{\nu} \mathrm{d} \nu$ by considering the following identity (Doppler effect):
\begin{equation}
\dfrac{dv}{c} = \dfrac{\mathrm{d} \nu}{\nu}
\end{equation}
In this case we have:
\begin{equation}
\int T dv = \dfrac{\lambda ^ 3}{2 k} \int I_{\nu} \mathrm{d} \nu
\end{equation}
Where $I_{\nu}$ is integrated over the frequency, and $T$ over (Doppler) velocities.
| {
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Will a free neutron radiate if it is decelerated? In this answer it is said (and I fully agree):
Yes, a ... photon can accelerate a lone neutron. The kinetic energy imparted to the neutron reduces the photon's wavelength (redshifts it) by the same amount, so the total energy of the system remains the same.
In turn, the opposite process has to be possible too. Neutrons are able to radiate. This usually is said only for charged particles.
Electromagnetic waves are emitted by electrically charged particles undergoing acceleration, and these waves can subsequently interact with other charged particles Electromagnetic radiation
Will a free neutron radiate if it is decelerated?
| We take a charged ball and shake it. There must be electromagnetic radiation from the shake. When we shake an electrically neutral object, it does not emit electromagnetic radiation from the shaking.
This situation cannot be explained by photons. Otherwise, an electrically neutral object will radiate when it is shaken. Because charged particles (proton, electrons) inside object radiate photons, objects radiate. But this is clearly not the case.
| {
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Proving that motion of an $n$ dimensional oscillator can be written as a linear combination of "sine waves" Here is a related question which might provide some context: LINK.
Let's consider an oscillator with equation of motion in $n$ dimensions:
$$
\frac{d^2}{dt^2} \vec{x} = K \vec{x}.
$$
Given that $\vec x=0$ is a stable equilibrium, how can I show that the system will oscillate in sine waves? In other words, how to prove that the system will not behave like $x=\sinh t$? More precisely, how can I show that the $y_i$ in the answer in the link above will be sine waves?
As the answer in the link above suggest, to prove the solution to the DE has sine wave pattern, I need to prove that $K$ is symmetric with negative eigenvalues (see also the comment below the answer).
What about the case that $\frac{d^2}{dt^2} \vec{x} = f(\vec{x})$, where $f(\vec{x})$ doesn't have to be linear, but can be approximated linearly by $K \vec{x}$? If $\vec x=0$ is a stable equilibrium, must the eigenvalues of $K$ be negative OR zero?
I hope I am clear. Please tell me if I am not expressing myself clealy.
| In the case that $K\vec{x}$ is exact rather than an approximation, note that if the equilibrium is stable, work done of a virtual displacement $\delta \mathbf x$ must be negative. So,
$$
\text{work done}=F.d\propto (K\delta \mathbf x).\delta \mathbf x=\delta \mathbf x^T K\delta \mathbf x<0.
$$
So $K$ is negative definite, and all eigenvalues are negative.
If $K$ is just an approximation of $f$, then we may only have $\leq$ instead of $<$. So you can still show that it is non-positive.
But why $K$ must be symmetric?
| {
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In a circuit having one resistor why do the electrons lose all their potential energy across that resistor and not do so if there are many resistors In a simple circuit which consists of a battery and one resistor, why do electrons lose all their potential energy across this one resistor regardless of the magnitude of the voltage or the resistance? Why is the voltage drop of this one resistor equal the voltage of the battery?
If there is more than just one resistor in series then in the first resistor happens a voltage drop which is not equal to the voltage of the battery. The electrons don't lose all their potential energy across the first voltage. it seems that the resistor knows that there is another resistor after it so it doesn't eat all the electrons' potential and leaves some potential for the next one. Why does this happen?
| It's simply due to energy conservation.
What does a battery? It uses its own internal chemical stored energy to create potential difference across its terminals and when it's terminals connected to the circuit it potential energy goes to electrons and if there is a resistor attached to the circuit.
then first what resistors do in the circuit? They dissipate the energy of electron which comes from a potential difference of battery terminals.
Hence if there only loss of energy can happen by a resistor is that circuit is isolated then energy provided battery is equal to the energy dissipated by resistor for which potential drop across the total resistance is equal to the potential difference across battery terminals.
It's nothing but KIRCHOFF'S LAW.
| {
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"url": "https://physics.stackexchange.com/questions/490109",
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Can there be interference between a proton and an electron? For example, we know that we can interfere two different electrons or two different protons by employing them in a double-slit experiment.
Now suppose, we mix protons and electrons and shoot them simultaneously for a double -slit experiment. Will the protons and electrons interfere with each other?
| The root of this confusion is the visualization that is very commonly given to explain wave particle duality. When we say that particles behave like waves, we do not mean that in the sense of a wave in a puddle. In quantum mechanics (and field theory), we say we have an electron wavefunction (in QFT, we say we have an electron wave). This wavefunction, when expressed in the spacetime basis ($x,y,z,t$), gives us the probability of finding an electron in a region in a time interval. Similarly, the proton wave has its own wavefunction.
But these two wavefunctions cannot be summed up to give us the joint probability distribution of the electron-proton pair. The waves in the puddle are representations of classical forces acting on a shared water surface and thus can be added (as vectors). But the electron wave and proton wave do not share any such surface. They are abstract representations of probability functions of two different quantum systems with their own characteristics and thus cannot be added that way --hence no interference.
To properly model the interaction between protons and electrons (or indeed two electrons), one needs Quantum Electrodynamics.
| {
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Newton's third law in magnetic fields Say I have a charged particle moving through a magnetic field perpendicular to it. It will experience a force, but according to Newton third law
Every force has an equal and opposite reaction.
So what is the opposite reaction/force of this magnetic force.
Which body experiences this force?
| The answer by user224659 is correct but I want to focus on the second part and I hope bring some added clarity.
The magnetic force is an interaction between a charged particle and an electromagnetic field. The rate of change of the momentum of the charged particle is
$$
\frac{d{\bf p}}{dt} = q {\bf v} \times {\bf B}
$$
and the rate of change of momentum in the local electromagnetic field (at the location of the particle) is
$$
\frac{d{\bf p}_{\rm em}}{dt} = -q {\bf v} \times {\bf B}.
$$
There you have an example of an "action and reaction" pair, to use the terminology of Newton's third law, but writing it in terms of rate of change of momentum might make it easier to see what is happening to the electromagnetic field.
Notice that when the charge is accelerating, the electromagnetic field is also changing: its momentum is changing and therefore it is not completely static. This comes about because there is both the applied magnetic field (which is owing to other things, not the charge under discussion) and also the electric field caused by the charge we are thinking about. The combination of these leads to the changing momentum in the electromagnetic field. (Keep in mind that
in order to carry a non-zero momentum, the electromagnetic field does not necessarily have to be in a wavelike motion.)
When we apply the above facts in practice, we don't treat point particles but rather local concentrations of charge, with a finite amount of charge per unit volume, and then $q$ refers to the total charge of such a body. The formulae above apply when the radius of the charged body is small compared to other relevant distances in the physical situation, but not so small as to yield unphysical predictions for the field very near the charge.
| {
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Why is dry soil hydrophobic? Bad gardener paradox When I forget to water my plants, and their soil becomes very dry, during the next watering I can see that the soil becomes hydrophobic. I can even see pockets of air between the repelled blob of water and the soil.
On the contrary, when the soil is moist, it very quickly absorbs the water.
This goes against the intuition, that the diffusion , in order to equalize the concentration soil-water, would create the opposite effect.
What is the physical explanation of this phenomenon?
I tried to search on google, but the explanations are of low quality and very "high level" without real physics involved.
| Diffusion and adhesion are different phenomena. Diffusion happens due to a concentration gradient, and this is how the water is slowly absorbed by the soil. This happens faster in dry soil per unit area per unit time than damp soil. On the other hand, adhesion is the ability of one material to stick to the other. Damp soil is more sticky to water than dry soil.
The main idea is that dry soil doesn't really stick to the water drops. The surface area is limited. But, damp soil literally gets surrounded by the water. This difference between surface areas result in damp soil absorbing water faster even though the process is slower per unit area.
| {
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Is $X\otimes X$ not the simultaneous position operator? I had thought that $X\otimes X$ would be the operator on $H_1\otimes H_2$ to simultaneously measure the x-positions of two particles.
But there seems to be something wrong with this -- for a given eigenvalue $z$, there is an entire subspace $\mathrm{Span}\left(|x\rangle\otimes|z/x\rangle\right)$ associated with it, so we don't get precise positions from measuring it, just the product of positions $x_1x_2$.
If so, what's the right operator representing "simultaneous measurement" of the x-positions? Is that even possible -- to have "vector eigenvalues"? Or do we just need a spacefilling curve or something?
| Let me first point out that this is not a conceptual issue but a "quirk" in the old formalism of quantum theory. Ask yourself this: if real numbers $x$ are the eigenvalues of position operator of a single particle then how the eigenvalues of a position operator of two particles should look like? If your answer is somethin like that $(x_1,x_2)$ then how do you write an operator with eigenvalues that are an ordered pair of 2 real numbers?
In the modern formalism we stop thinking about measurements in terms of eigenvalues and start thinking about them in terms of outcomes. Distinct outcomes come from distinct eigenspaces of the operator. The eigenvalues themselves are only labels that you may or may not define. In your case if $P_x^1\otimes I$ and $I\otimes P_x^2$ are the projections on eigenspaces of position for each particle then $P_{xy}^{12}:=P_x^1\otimes P_y^2$ are the projections on eigenspaces of position for both particles. The observable can be defined as $$X^{12}=\sum_{xy}\lambda_{xy} P_x^1P_y^2$$ where you can choose the eigenvalues $\lambda_{xy}$ however you want as long as they are distinct for all $xy$. The point is that the observable is not important, its the projections that are important.
A different approach is to think about it operationally. These are two observables that commute with each other. In the lab you would measure one particle, write down the outcome $x_1$ and immediately measure the second and write down the outcome $x_2$. Then you have the pair $(x_1,x_2)$ and you would say I have measured the simultaneous position of two particles. This suggests that if you have multiple commuting observables you don't have to come up with a whole new observable that measures both of them, you can just say I measure the set of commuting observables $\{X_1, X_2,X_3,...\}$
| {
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The solution to the non-linear convection equation The non-linear convection equation
$$u_{t} +uu_{x}=0$$ admits implicit solutions of the form $$u=f(x-ut).$$
How does one interpret this solution intuitively? Is there an example of a solution of this equation that can be written in an explicit form? How does an initial function evolve over time?
| *
*This quasi-linear 1st-order PDE
$$\left(\frac{\partial u}{\partial t}\right)_{\!x} + u\left(\frac{\partial u}{\partial x}\right)_{\!t}~=~0\tag{1}$$
is the inviscid Burgers' equation.
*It can be solved via the method of characteristics. The ODE IVP
$$ \frac{dx}{dt}~=~u, \qquad x(t\!=\!0)~=~\xi, \tag{2}$$
(where $u$ is treated as an external parameter and $\xi$ is an initial value) has integral curves
$$ x(t) ~=~ u t + \xi. \tag{3}$$
The initial value $\xi$ and the external parameter $u$ label the integral curves (3) in an $(t,x)$-diagram.
*A solution $u(x,t)$ to the PDE is constant along above integral curves (3):
$$ \frac{du(x(t),t)}{dt}~=~\left(\frac{\partial u}{\partial t}\right)_{\!x} + \left(\frac{\partial u}{\partial x}\right)_{\!t}\frac{dx}{dt}~\stackrel{(2)}{=}~ \left(\frac{\partial u}{\partial t}\right)_{\!x} + u\left(\frac{\partial u}{\partial x}\right)_{\!t}~\stackrel{(1)}{=}~0.\tag{4}$$
This leads to that the solution $u=f(\xi)$ should be a function of $\xi$.
*The function
$$x\quad \mapsto\quad f(x)~=~u(x,t\!=\!0)\tag{5}$$
is the initial profile.
*The implicit solution
$$u(x,t)~=~f(x-t u(x,t))~=~f(x-t f(x-t u(x,t)))~=~\ldots\tag{6}$$
is a fixed-point equation.
*Interestingly, the non-linear PDE can develop shock-wave singular solution $u$ from a smooth regular initial profile $f$.
*Example. The Wikipedia page lists the full solution
$$u(x,t)~=~\frac{ax+b}{at+1}\tag{7}$$ for the initial profile
$$f(x)~=~ax+b.\tag{8}$$
| {
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Abelian and non-Abelian holonomies I read the article Geometric Manipulation of Trapped Ions for Quantum Computation, and it mentioned “Abelian and non-Abelian geometric operations (holonomies)”. I know what is holonomy, and what is Abelian, but I didn't understand what are Abelian/non-Abelian holonomies.
I tried reading the references and searching Google, and found articles about those (non) Abelian holonomies, but with no explanations of what it is.
| A "non-Abelian holonomy" is the holonomy of a principal connection with the group of the principal bundle being non-Abelian. (In more physics-y parlance, it's the Wilson lines of a non-Abelian gauge theory)
| {
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How does alternating current provide energy? In my head, direct current makes complete sense; the electrons carry energy around the circuit to something being powered losing its potential and then return to the battery or whatnot to have their potential raised again.
This is probably wrong, so I would like an explanation of how a direct current actually transfers energy and also how an alternating current does (I have zero intuition for this).
| Here is an analogy which might help.
Imagine you are cutting a board with a saw. You push the saw through the board, and it cuts. Then you lift the saw up out of the board, pull it back, set it back into the board, and push it through again, always in the same direction. Eventually, the board is cut. Or...
You can push the saw through the board, and then pull it back while in the board, push it forward again, then pull it back, and so forth. As long as the saw teeth are shaped to cut on either stroke, you'll cut the board by going back and forth instead of in one direction only.
Running the saw one way only through the board is like DC power. Pulling the saw back and forth is like AC power.
| {
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Do air conditioner power ratings violate the 2nd law of thermodynamics? I just got a new AC rated at 6000 BTU and wanted to determine its power consumption. Some research on AC conventions quickly reveals that 6000 BTU really means 6000 BTU/h, where BTU is a measure of energy (British thermal unit). This is supposedly the rate at which heat is pumped from inside to outside. By the second law of thermodynamics, the power consumed to run the device must be even greater.
A direct conversion gives 6000 BTU/hr = 1758 W, so we should expect a power consumption greater than 1758 W. But I was surprised to see on the side panel of the unit (in the electricity specs) a voltage of 115 V and current of 4.66 A, suggesting a power of
$$115\text{ V} \times4.66\text{ A} \approx536\text{ W}.$$
At first I thought I must be misenterpreting the side panel, so I checked the official specs on the manufacturer's website. They boast a higher current of 6.5 A, and claim the electrical "rated input" of the device is 700 W. Still way too small. How is this possible?
P.S., all the sources I could find about this via googling seemed to contain similarly impossible numbers, without mentioning this seeming violation of the second law.
| Never mind, I'm pretty sure my statement that the power required to operate an air conditioner (or any heat pump) must be greater than its rate of pumping heat is wrong. I.e., the second law doesn't forbid the coefficient of performance (heat pumped / energy consumed) from being greater than one.
| {
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Why do jet engines sound louder on the ground than inside the aircraft? Everyone is familiar with the whirring sound of jet engines when seeing an aircraft taking off from a nearby airport. It is distinctly very loud on the ground and one can hear it even when the airplane is miles away.
Although one can hear a 'white noise' like sound when inside an airplane, the engines don't sound very loud in spite of being just meters away from them. I understand that the cabin is well insulated from the outside, but I would expect to hear a similar whirring sound of the engines.
So what is the phenomenon that makes jet engines sound louder on earth compared to inside the aircraft cabin?
| First, the cabin is quieter because the fuselage walls are designed to limit the transmission of sound from the engines.
Second, on most commercial aircraft, the engines are suspended beneath the wings, which block the noise from the engines before it can strike the fuselage.
| {
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Common potential in Capacitors If two isolated charged capacitors (of different capacitance) are connected in parallel to each other they acquire a common potential. But suppose if i connect positive plate of one capacitor to negative plate of another capacitor will they still acquire a common potential or will the charge acquired by two capacitors be same as the circuit looks like that the capacitors are connected in series in which -ve plate of capacitors is connected to + ve plate of another capacitors
| If positive plate is connected to negative still the process remains the same charge will flow untill two capacitor are at same potential.
| {
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Intensity mapping of the 21 cm line I'm currently reading Astrophysics for physicists by A.R. Choudhuri. Section 6.5 of his book he presents the following plot:
and tries to explain how it was generated. I'm having a hard time understanding how we can differentiate between the different velocities. As far as I understand we only measure the intensity $I(l,0)$ along the line of sight for some $l$. So I would expect that we are able to plot $I(l,0) \text{ vs. } l$, but I don't see how we can actually know how the velocity distribution of the ISM along the line of sight is.
In his book, Choudhuri talks shortly about
$$v_R = (\omega-\omega_0)R_0\sin l,$$
where $\omega_0$ and $R_0$ are the angular speed and the radius of our sun to the galactic center. I think he uses this to somehow explain how we can actually measure $I(l,0,v_R)$ instead of $I(l,0)$ but I just couldn't follow his arguments.
| Actually the velocity can be measured by utilizing
the Doppler effect.
The frequency of the 21-centimeter hydrogen radiation
is known with high precision: $f_0 = 1.420405752\, \mathrm{GHz}$.
We also know the speed of light: $c = 2.998 \cdot 10^8 \mathrm{m/s}$.
The Doppler effect is the following. When the radiating source is approaching you with velocity $v$,
then you measure a slightly higher frequency
$$f = \left(1+\frac{v}{c}\right) f_0.$$
Likewise, when the radiation source moves away from you,
then you measure a slightly lower frequency.
Hence, by measuring the frequency $f$, you can calculate the velocity
$$ v = \frac{f-f_0}{f_0} c.$$
| {
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What is the difference between uniform velocity and constant velocity? I think that uniform velocity implies constant speed but not constant direction. while constant velocity implies constant speed without any changes in direction.
Both tell us that there's no acceleration (since magnitude of velocity is constant).
The same goes for acceleration: both imply constant magnitude, but only constant acceleration means that there's no change in its direction.
However, a lot of people on the Internet argue that whether it's the other way around or that there's no difference at all. Who's right and who's wrong?
| Regardless of what you call it, in order for an object to not be accelerating both the magnitude of its velocity (speed) has to be constant AND its direction (path) needs to be in a straight line (aka rectilinear motion).
In order for the direction of an object to change it must experience a net force and thus an acceleration.
Take the simple case of an object moving in a circular path at constant speed. It experiences a centripetal acceleration of magnitude
$$a=\frac {v^2}{r}$$
And a force of
$$F=m\frac {v^2}{r}$$
Hope this helps
| {
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Potential and potential energy I know when a negative charge moves in the direction of a uniform electric field its potential energy increases and its potential decreases. For example, its potential energy changes from $0.9\ \rm{mJ}$ to $1.2\ \rm{mJ}$, but its potential changes from $-90\ \rm V$ to $-120\ \rm V$. Where is the potential zero, and where is the potential energy zero?
Do they become zero at different locations?
| Electric potential is just the electric potential energy per charge. In other words, electric potential just depends on the charge distribution around you, where as if you were looking at a charge in the field caused by the charge distribution you could then say it has an associated potential energy in that configuration.
In light of this, the relation between electric potential energy $U$ and electric potential $V$ for a charge $q$ is just $$U=qV$$ Therefore, these two values need to be equal to $0$ at the same point in space. Of course, this $0$ point can be chosen to be at various locations, but once you set it then you have to stay consistent, and $U$ and $V$ will both be $0$ at that location.
Therefore your example is totally fine. If the potential energy goes from $0.9\ \rm{mJ}$ to $1.2\ \rm{mJ}$ and your potential goes from $-90\ \rm{V}$ to $-120\ \rm{V}$ nothing is wrong. In either case the values are moving farther away from $0$. If you moved the charge in the other direction back to where one value was $0$, you would find the other value to be $0$ as well.
| {
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EM-Wave: Calculate magnetic field $H$ from electric field $E$ In an exercise I am supposed to calculate the magnetic field from the electric field for a plane, harmonic wave in vacuum.
$$\vec{E} = - E_0 \cdot \sin(\omega t - k z) \cdot \vec{e_y}$$
Using the law of induction
$$\operatorname{rot} \vec{E} = -\mu \dfrac{\partial \vec{H}}{\partial t} $$
I end up with this solution for the $x$ component of $H$,
\begin{align}
\dfrac{\partial H_x}{\partial t} & = \dfrac{E_0 k}{\mu} \cos(\omega t - k z)\\
H_x(z,t)& = H_x(z,0) + \dfrac{E_0 k}{\mu \omega} (\sin(\omega t - k z) - \sin(-kz)).
\end{align}
According to the provided solution this is right, except for the integration constant $H_x(z,0)$. How do I choose this constant?
Intuitively I would use the wave impedance $Z$,
$$|\vec{H}| = \dfrac{1}{Z} \cdot |\vec{E}|,$$
but if I can choose the constant $H_x(z,0)$ as I want, this wave impedance formula seems to be false as well...
| As my2cts mentions in the comments, the magnetic field in the presence of currents is only partially determined by the electric field. However, in vacuum they do determine each other fully.
Remember that the magnetic field is not only governed by Farday's law of induction
$$ \epsilon_0^{-1}\mathrm{rot} D = -\mu_0\partial_t B $$
but also by Ampère's law
$$ \mathrm{rot} H = \partial_t D $$
You already have evaluated the former and found the solution up to a constant of integration $H(\vec{x},0)$. If you evaluate the latter as well, you will be able to fix $H(\vec{x},0)$ up to a true constant $\vec{C}$. Maxwell's equations do not forbid that solution. However, this background field needs to vanish, because space(time) is isotropic and translational invariant. A non-vanishing constant background would violate Galilei/Lorentz-invariance.
| {
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Is the only absolute difference between types of light frequency? Probably a bad question but for some reason, it seems too simple in my head that anyone at home could theoretically create anything from radio waves to gamma waves by generating electrical signals at different frequencies.
Say I had a electronic frequency generator that was able to produce a signal at any frequency, and for illustrative purposes, say there was a diode hooked up this generator that could receive its signals.
If it created a signal at $10^{12}$ Hz, the diode would give off infrared radiation.
If I increased the signal to $10^{20}$ Hz, the diode would give off gamma radiation.
I’m using this example just to emphasize my question, is frequency the absolute and only differentiator in types of light on the electromagnetic spectrum?
|
[F]or some reason, it seems too simple in my head that anyone at home could theoretically create anything from radio waves to gamma waves by generating electrical signals at different frequencies.
I don't see how to easily create a device that vibrates at 10¹² Hz, not even 10⁶ Hz. Technologically, this is not as simple as it sounds.
As mentioned by @Brick, the "electromagnetic spectrum" includes all frequencies. What makes a wave "electromagnetic" is not the frequency but the nature of the phenomenon. In electromagnetic waves you have oscillations of electric and magnetic fields through space, and also a transport of energy while the wave propagates.
| {
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Relativistic beaming/ Aberration effect derivation I'm researching applications of relativistic beaming and I want to derive a formula for the aberration effect but I am stuck (I am off by a factor of 1/c). Here's what I have:
Consider a star in its rest frame, $S^\prime$, moving at speed $v$ that emits a photon at an angle $\theta^\prime$. In the $S$ frame, the observer measures the angle to be $\theta$. I made a picture below to illustrate this
We have from the picture the identities $\cos(\theta)=\frac{x}{ct}$ and $\cos(\theta^\prime)=\frac{x^\prime}{ct^\prime}$. The Lorentz transformations are from the prime coordinate system are the following:
$$x^\prime=\frac{x-vt}{\sqrt(1-\beta^2)}=t\frac{\cos(\theta) -\beta}{\sqrt(1-\beta^2)}$$
$$y^\prime =y$$
$$t^\prime=\frac{t-\frac{vx}{c^2}}{\sqrt(1-\beta^2)}=t\frac{1-\beta\cos(\theta)}{\sqrt(1-\beta^2)}$$
Now, $$\cos(\theta^\prime)=\frac{x^\prime}{ct^\prime}=\frac{t\frac{\cos(\theta) -\beta}{\sqrt(1-\beta^2)}}{ct\frac{1-\beta\cos(\theta)}{\sqrt(1-\beta^2)}}$$
After simplifying we get $$\cos(\theta^\prime)=1/c \frac{\cos(\theta)-\beta}{1-\beta\cos(\theta)}$$
Which is almost the relativistic aberration formula but again I am off by $c^{-1}$. Does anyone know what is wrong with this derivation?
| $$\cos(\theta)=\frac{x}{ct}\rightarrow x=ct\cos(\theta)$$
$$c\beta=v$$
$$x'=\frac{x-vt}{\sqrt{1-\beta^2}}=\frac{ct \cos(\theta)-c\beta t}{\sqrt{1-\beta^2}}=ct(\frac{ \cos(\theta)-\beta }{\sqrt{1-\beta^2}})$$
| {
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Do photons exchange energy with objects according to their momentum? I've been thinking about the doppler effect, where electromagnetic radiation experiences a redshift if the radiation source is relatively moving away from the observation point, and a blueshift if it's moving closer.
If I have a light source, a sensor which is stationary relative to the light source, and a mirror either moving towards or away from the light source while reflecting the light to the sensor — the sensor would pick up the electromagnetic waves as shifted, no? And thus photons (if they are being reflected) basically lose energy to or gain energy from an object (with which they interact) in accordance to the object's momentum relative to the vector of the wave?
I believe that should be the case as I know solar sails are "pushed" by light. Thus I assume light in a way should be able to be "pushed back" and gain energy (which causes the frequency shift).
I'm sorry if this is an obvious concept already explained by the doppler effect. I'm not formally learned in physics and didn't find this specific phenomenon explained in the texts I read.
| The collision between an object and a photon can be modelled like any other collision in physics; you take a reference frame (I recommend the rest frame of the object being struck) and give the photon an energy $h\nu$ and a momentum $\frac{h\nu}{c}$. If the photon is absorbed, then the object gains velocity $\frac{h\nu}{M}$, or if it is reflected then the velocity $v$ and reflected/transmitted frequency $\nu'$ solve the equations $\frac{1}{2}Mv^2+\frac{h\nu'}{c}=\frac{h\nu}{c}$ and $Mv\pm h\nu' = h\nu$. This can lead to the photon losing energy and increasing its wavelength.
| {
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Deriving Current density of a Moving Point Charge Using the Continuity-Equation Problem:
We know, a point charge at position $\mathbf{r}_q$ has the charge density
$$\rho_q(\mathbf{r})=q\delta(\mathbf{r}-\mathbf{r}_q) \tag{2}$$
if it moves with the velocity $\mathbf{v}$, we get the current-density:
$$\mathbf{j}_q(\mathbf{r})=q\mathbf{v}\delta(\mathbf{r}-\mathbf{r}_q) \tag{3}$$
Apparently, we can derive (3) using the continuity equation
$$\dot{\rho}+\text{div} \ \mathbf{j} = 0 \tag{4}$$
and the fact that
$$\mathbf{v}=\dot{\mathbf{r}} \tag{5}$$
Question:
I really can't derive (3) using (4). At some point, I'd have to write down $\dot{\rho}$ no? But I have no idea how to evaluate that. Also I have no idea how to get rid of the "div". I just don't see how to even start here.
| Note that $\mathbf{r}_q$ is a function of time, and thus so is $\rho_q$. We have
$$ \rho_q = q\delta(\mathbf{r} - \mathbf{r}_q(t)) $$
You can use the identity
$$\frac{\partial}{\partial t}f\left(\mathbf{A}(t)\right) = \frac{\partial\mathbf{A}}{\partial t} · \mathbf{\nabla}f(\mathbf{A}(t)) $$
to find
$$ \dot{\rho} = -q\mathbf{v}_q · \mathbf{\nabla}\delta(\mathbf{r} - \mathbf{r}_q). $$
Assuming $ \mathbf{v}_q $ does not change, using the identity
$$ \mathbf{\nabla} · (\phi\mathbf{A}) = \mathbf{\nabla}\phi ·\mathbf{A} + \phi\mathbf{\nabla}·\mathbf{A} $$
we can write
$$ \dot{\rho}=-q\nabla · \left( \mathbf{v}_q \delta(\mathbf{r} - \mathbf{r}_q) \right)$$
and by the continuity equation,
$$ \nabla · \mathbf{j} = -\dot{\rho} = q\mathbf{\nabla}·\left( \mathbf{v}_q \delta(\mathbf{r} - \mathbf{r}_q) \right)$$
so we have
$$ \mathbf{j} = q\mathbf{v}_q \delta(\mathbf{r} - \mathbf{r}_q). $$
This is not the only solution, however. Any vector field that differs from $\mathbf{j}$ by a divergence-free vector field also satisfies the continuity equation. This is all the continuity equation can tell us, as it provides no information on $ \mathbf{\nabla} \times \mathbf{j} $.
Quite generally, for a charge distribution $\rho$, we have $\mathbf{j}=\mathbf{\rho}\mathbf{v}$, where $\mathbf{v}$ is the velocity field of this charge distribution. This is clearly also true if you accept the expression for $\mathbf{j}$ above, with no additive divergence-free vector field.
| {
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Does magnetic moment change under inversion symmetry? Since magnetic moment can be view as a small electric current circle. Pictorially, when apply inversion operation, the current direction is reversed, so I think the $\vec{m}\to -\vec{m}$ under inversion symmetry operation.
On the other hand, the formula for the magnetic moment is $\vec{m}=\int_V \vec{r}\times\vec{j}\mathrm{d}V$. Under the inversion symmetry operation $\vec{r}\to -\vec{r}$ and $\vec{j}\to -\vec{j}$, therefore $\vec{m}$ is unchanged.
The above two reasoning must have one being wrong, which one and what is the flaw of the reasoning?
| $\vec m$ and $\vec B$ are indeed "axial vectors" and invariant under space inversion. Imagine a circular current. Under inversion the current changes direction but also the circle is inverted. This leaves $\vec m$ unchanged. (just noticed Ben Crowell's ear lier answer which is identical to mine.
$\vec E$ also is not a true 3D vector as it changes sign under time reversal.
| {
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Motion between two particles in a relative manner Suppose a particle A is travelling in east direction with velocity of x m/s and another particle B is travelling with velocity y m/s in the west direction. Why does the the particle B appears to move towards A with a velocity of x+y and not just y m/s?
|
given two vectors $\vec{v}_{01}$ and $\vec{v}_{02}$
thus:
$$\vec{v}_{12}=\vec{v}_{10}+\vec{v}_{02}\,,\quad\text{("zero cancel")}$$
where
$\vec{v}_{10}=-\vec{v}_{01}$
your example
$\vec{v}_{01}=x$
$\vec{v}_{02}=-y$
$\Rightarrow$
$\vec{v}_{12}=-x-y=-(x+y)\quad \surd$
$\vec{v}_{21}=-\vec{v}_{12}=(x+y)$
| {
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Why does an inductor oppose the change in current (magnetic field)? May I get a physical interpretation on this question? What is happening in the inductor when the current is running through it and what is physically happening when the current starts changing?
| Here is one way of looking at this.
We start with an inductor that has a steady current flowing through it from a power source. Because of this, there is a magnetic field extending into space surrounding the inductor.
Now we attempt to cut off the flow of current through the inductor, by switching off the source. At the instant the current goes off, the magnetic field begins to collapse around the inductor, which induces a current flow in the inductor in the same direction as our original current. The quicker the field collapses, the greater the induced current flow- and we observe a big fat spark jumping across the switch terminals as they move apart.
We conclude that the current flow in the inductor really wants to keep flowing, and the inductor "fights" any change in the magnitude of the current flow we try to assert.
| {
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Resistors are used to reduce current to prevent light bulbs from "exploding" but it's also said that "current remains same at all points in a circuit" Resistors are used to reduce current in order to prevent light bulbs and other electrical components from “exploding”, but it is also said that “current remains the same at all points in a series circuit”.
Then what's the point of using resistor when the current will remain the same, and as it would remain the same, the light bulb would still receive that high current which could cause it to explode.
So, again, why are we using resistors? If current is still going to be the same, what's the point? The light bulb would still explode if it will not be receiving a lower current. Also, I'm talking about the series circuit, not a parallel one.
| You misunderstood the phrase "current remains constant in a series circuit".
Consider the circuit below with a light-bulb and a resistor in series.
In this circuit the current through the light-bulb is the same
as the current through the resistor.
This is meant by "the current remains constant in a series circuit".
The current according to Ohm's law is:
$$I = \frac{V_{\text{battery}}}{R_{\text{bulb}}+R_{\text{resistor}}}$$
Notice that this current is the same all the way around the circuit,
because the current cannot leave the wire anywhere.
Thus the electrical current behaves very much like the water current
flowing through a pipe.
The main difference between both is: The electrical current is
measured in Ampere, and the water current is measured in liter/second.
Now consider the next circuit with a light-bulb only.
In this second circuit the current through the light-bulb is greater
than the current through the light-bulb in the first circuit.
This is because the total resistance (light-bulb only) is lower
than in the first circuit (light-bulb + resistor).
Now the current according to Ohm's law is:
$$I = \frac{V_{\text{battery}}}{R_{\text{bulb}}}$$
which is greater than the current in the first circuit.
| {
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Does the law of conservation of momentum mean both linear and angular momentum? I was solving a problem where a bullet hit a rod hinged about one of its ends. The rod is standing vertically before the collision. We had to find the final angular velocity of the rod. The way we did it was to find the angular momentum of the bullet about the axis of rotation just at the point where it touches the rod.
I have two questions.
*
*Firstly, I thought that momentum was always conserved. So when the bullet approaches the rod the total momentum of the bullet is its linear momentum, which must be equal to the total momentum of the combined system after the collision(which now becomes angular). Where am I going wrong with my logic?
*Secondly, I am confused about the idea of having to find the angular momentum for an object that is traveling linearly. What does this mean physically?
| The angular momentum conservation law is independent of the linear momentum conservation law. It is two laws, that means that they have been established by experiment, together with energy conservation that are always fulfilled in the mathematical theories we at presently have for physics, both quantum and classical.( It is only in General Relativity one has to think twice about these experimentally established laws , in special cases).
Once a mass has linear momentum $p$ an angular momentum , $L=r\times p$, can always be defined for any point in space with respect to this particle, in effect assuming that there is an axis of rotation at that point ( it is as if the particle is flying off tangentially to the hypothetical axis).
Now in the specific problem, the bullet comes with a calculable angular momentum towards the axis around which the rod will rotate.
The conservation of linear momentum is taken care by the whole system ( earth + rod), since the mass of the earth is so large it can be assumed for the system to be immobile, if the rod were not well attached to the earth, the whole thing would also move to conserve linear momentum.
The angular momentum that the bullet had is conserved by the rotation of the rod, and the energy for the whole operation is given by the energy brought in by the bullet.
| {
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Spacetime - reasons for "unification" of space and time - why we imply some unity in the term spacetime? Why do we say spacetime, as though stressing that this is some kind of unity: space together with time?
I understand a little that mathematically we have 4 variables in the same (Einstein) equation(s) (3 for space, 1 for time), but can it be put in other words?
P.S. This question was inspired buy reading quantum origin for spacetime review (idea that entanglement produces space-time) and especially the original paper but it is another story - matter sources curvature (at macro level) and matter can be in a superposition of different positions (at quantum level), allowing superpositions of different curvatures. In other words, "quantum gravity" allows superpositions of different geometries.
|
Why do we say spacetime, as though stressing that this is some kind of
unity: space together with time?
Special Theory of Relativity (STR) teaches that time is a coordinate rather than a universal parameter.
That is, when we transform (using the Lorentz transformation ) from one inertial coordinate system (ICS) to another, relatively moving ICS, the time coordinate is involved in this transformation so that, for example, a spatial coordinate in one ICS is a mixture of one or more spatial coordinates and the time coordinate in the other ICS.
That is, space and time are now 'woven' together into a spacetime where a purely spatial or temporal translation in one ICS is a mix of both temporal and spatial translations in a relatively moving ICS.
| {
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Sequential Stern-Gerlach experiment Consider the following diagram:
(Sakurai)
In the first lecture of MIT OCW Quantum Physics 1, 2013 (https://www.youtube.com/watch?v=lZ3bPUKo5zc), Allan Adams implies that if we remove the barrier on the Sx- beam, thus allowing the Sx+ beam and Sx- beam to merge together and pass through the next SG measurement setup, the output of that final measurement will be only Sz+ atoms, rather than both Sz+ and Sz-. Why does this happen? Why do the atoms not care about being collapsed into Sx eigenstates in this scenario?
| Consider an initial state
$$|1 \rangle = \frac{1}{\sqrt{2}} (|S_z + \rangle + |S_z -\rangle). $$
Now it passes through the first SG experiment, and all down spins are filtered out, so you have a collapsed state
$$|2 \rangle = |S_z + \rangle = \frac{1}{\sqrt{2}} (|S_x + \rangle + |S_x -\rangle). $$
If a measurement is made after the second SG experiment, you again collapse into a state $S_x+$
$$|3 \rangle = \frac{1}{\sqrt{2}} (|S_z +\rangle + |S_z - \rangle) = |1\rangle. $$
However if no measurement is made, and the beams are merged, your system is still described by $|2 \rangle$, so a measurement of $S_z$ will only yield $|S_z +\rangle$.
The key point is that splitting out into the two $S_x$ components and merging again without ever measuring will not collapse the state. Superposition can persist over distance as long as no measurement is made, see e.g. the quantum erasure experiment with spacelike separation for practical examples of this.
For reference, the spin states are related as describe on Wikipedia.
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Causality under relativity If simultaneity is relative; meaning that for each any events A and B and we can find a reference frame in which two events A and B occur in inverse order, how does our notion of 'causality' withholds?
| Spacetime causality holds within the light cone. Only events within the past light cone can affect the present event (the here-and-now at the vertex of the light cone). The present event can affect only the future light cone. Events outside the light cone can neither affect nor be affected by the present event.
Lorentz transformations can change the temporal ordering of events outside the light cone relative to the present event, but they cannot change the temporal ordering of events inside the light cone relative to the present event. All events in the past light cone are temporally before the present event in all Lorentz frames, and all events in the future light cone are temporally after the present event in all Lorentz frames.
The essence of the events inside the light cone are that they are closer in space than they are in time to the present event, so signals traveling at the speed of light or less can travel from those in the past light cone to the present event or from the present event to those in the future light cone. To get to or from events outside the light cone would require traveling faster than light.
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What's wrong with this argument that the potential energy of an arbitrarily heavy object at arbitrary height is $0$? Consider an object with mass such that there is a gravitational force downward of $1N$. Also assume the environment is a perfect vacuum. Now assume that we exert a force of $1+\epsilon $ Newton upward on the object for $\delta$ seconds, and then exert a force of $1$ Newton upward from then on.
During the first $\delta$ seconds, the object will accelerate from $0$ to $v$ meters per second upward, and then stay at $v$ since our force balances the gravitational force. So the object will move upwards indefinitely.
During the first $\delta$ seconds, we have a net upward force of $\epsilon$ Newton, and so a work done on the object of $\epsilon \cdot \frac 1 2 v\delta\approx 0$ (the force times displacement). After the $\delta$ seconds, the net force is $0$, since the gravitational force and our force cancel out, and so the net work done on the object is $0$, so the object doesn't gain any energy even though its height increases.
What's wrong with this argument?
| The point is that when a mass is not moving in a gravitational field, the amount of potential energy it has depends on its position.
If you release a mass 1m above the ground and it falls freely, it has less kinetic energy when it hits the ground than if your released it from 2m above the ground.
That difference in energy is what "gravitational potential energy" measures.
If you raise the mass at constant velocity, the force you apply is doing work (= force x distance), the kinetic energy stays the same, and the gravitational potential energy is also increasing.
Of course the gravitational force is also doing (negative) work equal and opposite to the force you apply, but "the work done by gravitational force when a mass changes position" and "gravitational potential energy" are just two names for the same thing (though numerically they have equal and opposite signs).
| {
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How can we tell if the Earth is spinning without any external references? The rotation of the Earth about its axis makes it bulge at the equator and contract at the poles due to the centrifugal forces. How do we know, without any external references, that the Earth is spinning if there is nothing to compare it to? For example, imagine it was spinning in empty space with no other objects. Does it still bulge at the equator?
Particularly, what can we say theoretically about it?
| Earth rotation can be detected using a gyroscope (http://www.tkt.cs.tut.fi/research/nappo_files/Symposium_Gyro_Technology_2010_web.pdf). It looks like the equipment used in the article costs under $1000 (not including a computer).
EDIT (8/24/2019): Another approach was proposed by Compton (Science 23 May 1913:
Vol. 37, Issue 960, pp. 803-806):
"if a circular tube filled with water is placed in a plane perpendicular to the earth's axis, the upper part of the tube with the water in it is moving toward the east with respect to the lower part. If the tube is quickly rotated through 180 degrees about its east and west diameter as an axis, the part of the tube which was on the upper side attains a relatively westward motion as it is turned downwards (since it is drawing nearer the earth's axis). But the water in this part of the tube retains a large part of its original eastward motion, and this can be detected by suitable means."
"If then $\alpha$ is the angular velocity of the earth's rotation, $r$ the radius of the circle into which the tube is bent..., the relative velocity between the water and the tube when it is quickly turned from a position perpendicular to the earth's axis through 180 degrees is...$\alpha r$."
"In order to prevent convection currents, it is best to hold the ring normally in a horizontal position, in which case the relative motion is of course $\alpha r \sin\phi$, where $\phi$ is the latitude of the experimenter."
For $r=0.993 m$ and the latitude $40^{\circ} 48'33''$ (Wooster, Ohio), we obtain the velocity of about $0.05mm/s$; the average velocity observed by Compton agreed with the calculated velocity with 5% accuracy. Compton measured the water velocity by observing globules of oil introduced in the water using a micrometer microscope.
| {
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Why does $k_\text{B} T \ll \hbar\sqrt{k_\text{Hooke}/m}$ imply that vibrational motion is negligible? I want to estimate the heat capacity of a diatomic molecule whose movement has been constrained to a 2-dimensional plane.
I can assume that
$$
k_\text{B} T ~\ll~ \hbar\sqrt{\frac{k_\text{Hooke}}{m}}
\tag{1}
\,,$$ where $k_\text{Hooke}$ is the Hooke's law constant.
The answer is $\frac{3}{2} k_\text{B} ,$ since there are 2 translational movements and one rotational, so:
$$
U = \frac{f}{2}k_\text{B} T
~~\implies~~
C = \frac{3}{2} k_\text{B}
\,.$$
Since the heat capacity, $C ,$ isn't dependent on $k_\text{Hooke} ,$ this seems to say that vibration can be neglected. However, how does the assumption in $\operatorname{Eq.}{\left(1\right)}$ end up resulting in this conclusion that vibration can be neglected?
Question: How does $k_\text{B} T \ll \hbar \sqrt{k_\text{Hooke}/m}$ imply that the vibrational motion is frozen out?
| To help understand what your assumption is saying, it might be easiest to start by defining what each side measures. What does $k_B T$ measure? What does $\hbar\sqrt{k/m}$ measure? What are the units of them?
And once you figure out the units and what each side represents, what does it mean that one side is much, much less than the other side? What would happen if the inequality is reversed? What happens when they are of similar order of magnitude?
| {
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Must an operator that preserves probability be unitary? One property of the unitary operator is that it preserves the norm of the state-vectors:
$$
\langle \Psi | U^\dagger U | \Psi \rangle = \langle \Psi | \Psi \rangle
$$
If $U$ is unitary.
Is the inverse statement also true? For an operator that will satisfy above equation for all $|\Psi \rangle$, will that operator be unitary?
| If $U:H\to H$, with $H$ a Hilbert space, is linear it turns out that it preserves the norm (in particular it is injective) if and only if it preserves the scalar product. The proof is based on the so-called polarization identity as Emilio suggested. However this does not mean that the operator is unitary, since the condition $UU^*=I$, corresponding to surjectivity, is not necessarily valid if the space has not finite dimensione. If the dimension is finite, injectivity and surjectivity are equivalent as is known from elementary linear algebra.
| {
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What prevents ice from being an electret? My question is apparently simple: if we put water in a electrostatic field and leave it to freeze, while still in the strong electrostatic field, to make ice, why wouldn't this ice exhibit electret capabilities?
| In the structure of ordinary hexagonal ice at low pressure, each molecule is held in a sort of tetrahedral cage by its four nearest neighbors. It acts as donor to two hydrogen bonds, and acceptor to two more. You might get the wrong idea that it can have any of $\left( _{2}^{4} \right)=6$ possible orientations, but the orientation of each molecule is constrained by the orientations of its neighbors. (There is some freedom, but flipping one molecule induces a cascade of flips.) In the absence of an applied electric field, ice has no overall dipole moment.
An applied field would, of course, induce enough polarization to give ice its impressive dielectric constant, but the polarization is not frozen in. If the applied field is turned off, the molecular orientations relax in roughly a microsecond, as inferred from the frequency-dependent dielectric constant. (The relaxation time is roughly $10^4$ times shorter in liquid water.)
It is not unreasonable to suspect that an applied electric field might favor one of the alternative structures observed at extreme pressures, which have less orientational entropy, but don’t forget that the overall dipole moments of electrets get neutralized by surface charges attracted from the environment.
| {
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Are the electrons' orbitals the same for all atoms? Are the electronic orbitals of an atom always quantified in the same way (i.e. the same energy required to reach the next level), or does each atom have its own values for each level?
If the quantification is universal, then the creation of photons (due to the deexcitation of the electrons) at the wavelength / color corresponding to the transition should be more abundant in the universe than all the other frequency. Except one detects no more photon of a given wavelength than of another.
So where is my reasoning error?
| The energy levels depend on two things:
*
*the electrostatic attraction between the electrons and the nucleus
*the electrostatic repulsion between the electrons
If you take a hydrogen atom, which is what your diagram shows, then there is a single electron and a single proton. The electron is attracted to the proton and there is no electron-electron repulsion because there is only one electron.
If you move on the the next element, helium, there are two electrons and the nucleus contains two protons. So the attraction between the electrons and the nucleus is now twice as big but we have a repulsion between the two electrons. Both these factors change the energy levels so they are not the same as hydrogen. The next element, Lithium, has three electrons and three protons in the nucleus so the energy levels are different again. And so on.
So all the atoms of a given element have the same energy levels because they have the same numbers of electrons and protons. For example all hydrogen atoms have the same energy levels. But the different elements have different energy levels because they contain different numbers of electrons and protons. The hydrogen energy levels differ from helium, which in turn differs from lithium and so on.
And just to complicate matters the number of neutrons in the nucleus makes a small difference as well, so for example the energy levels of hydrogen are slightly different from the energy levels of deuterium and tritium.
| {
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What's the point of dimensional regularization? I'm studying regularization of divergent integrals in QFT from Here: Roberto Soldati - Field Theory 2. Intermediate Quantum Field Theory (A Next-to-Basic Course for Primary Education)
I think I'm missing the big picture, let me explain.
At page 166 he says
In order to give a precise mathematical meaning to the above listed ill defined integral expressions, we have to introduce from the outset some kind of regularization procedure, the aim of which is to build up absolutely convergent loop integrals.
Bold mine.
Then pages later (from page 178), when he explains dimensional regularization he does it with a paradigmatic example, he considers this integral:
$$
I = \mu^{4 - 2\omega} \int \displaystyle\frac{d^{2\omega} k_E}{(2\pi)^{2\omega}} \big(k_E^2 - \Delta \big)^{-2}
$$
We then obtain (page 181 equation 4.23)
$$
I = \frac{1}{16 \pi^2} \bigg( \displaystyle\frac{1}{\epsilon} - {\bf{C}} + ln \displaystyle\frac{4\pi\mu^2}{\Delta} \bigg) + O(\epsilon) \tag{4.23}
$$
Where $\epsilon = 2-\omega$.
Now if we want $I$ in 4 dimensions we have to take $\epsilon = 0$ and we see in $(4.23)$ that with this choice $I$ diverges because of the term $\displaystyle\frac{1}{\epsilon}$, therefore we didn't built a convergent integral as stated in the bolded part of the quote above. Then I ask:
What did we achieve with this procedure if when we come back to four dimension we still have a divergent result?
And since I am probably missing the point and the big picture please feel free to explain it to me.
| The integral is absolutely convergent for $\epsilon >0$. The point is the following. Frequently in QFT we care about an amplitude $A$ which can be expressed as the sum of multiple integrals
$$A = I_1 + I_2 + \cdots + I_n.$$
Each integral, roughly, corresponds to a different Feynman diagram. The sum $A$ is the only physical quantity we care about, and therefore it must be finite. For reasons having to do with both UV and IR divergences, each integral $I_i$ may be infinite, but in such a way that the infinities cancel out. This is the program of renormalization (for the UV divergences) and inclusive summation / KLN theorem (for the IR divergences).
So as you know we regularize each by (for instance) writing $d=4-\epsilon$, evaluating them all, summing them together, and then taking $\epsilon \to 0$ at the end. So indeed $\lim_{\epsilon\to 0} I_i$ may equal $\infty$ for any single integral, but if we have understood the physics correctly then $$\lim_{\epsilon\to 0} (I_1 + \cdots + I_n) = \text{finite.}$$
In other words the problematic $\frac{1}{\epsilon}$ (or higher power) terms cancel.
| {
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What happens if a traveling sound wave encounters vacuum? Suppose a sound wave is emitted by an object in a medium like a gas so it travels in some direction. If the wave meets a rigid object, for example a wall, it reflects back as one should expect; and if it encounters another medium, like a denser one, it will be transmitted to the other side. In those cases the sound wave keeps traveling, and of course it can loose energy in its path and even be absorbed by some object, so it can be converted into another form of energy.
Now, I'm aware that sound waves can't travel in vacuum, so my question is: What happens to a sound wave that is traveling in some medium and encounters vacuum ?. Here's a drawing of the situation:
I'm thinking about the usual wave phenomenoma and reflection is not a logical option since there is not a defined object that can work like a wall, and transmission doesn't make any sense at all since sound waves can't travel through vacuum. So what happens with that wave ? Where does its energy go ?.
| The Sound wave would continue to propagate outward until it cannot continue, even while in a vacuum, and like many of us know, it would not be considered sound unless there are particles to be pushed against it and received by us through our sensation of hearing, and a Vacuum is a place devoid of said particles to carry sound.
| {
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Current in RC circuit Why is current drawn in an RC circuit (in a circuit powered by DC voltage supply) independent of the capacitor used?
While the capacitor is charging current drawn from the battery only depends on resistor and on the value of capacitor.
| In a charging RC circuit where the capacitor is initially uncharged, the charges will move as if the capacitor is essentially absent. Therefore, the initial value of the current is just equal to $V/R$.
If the RC circuit starts with a fully charged capacitor and is discharging, then once the current starts the capacitor acts like a battery. The circuit is then essentially a resistor in series with a battery, and the initial current is once again $V/R$ (since the capacitor's initial potential difference has a magnitude of $V$).
At times when the current is changing, the value of the capacitance does effect how quickly the current changes with time constant $RC$. This is because the amount of charge stored on the capacitor changes the potential drop across the capacitor due to $V=CQ$. The more charge that is present, the more it “fights the battery” in the charging case, and the more it pushes charge of itself in the discharging case.
In either case, the magnitude of the current is given by
$$I(t)=\frac VRe^{-t/RC}$$
so the current does depend on the capacitance, but the initial current does not.
| {
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Can anyone explain the Planck area? First of all, I am not an expert. I was reading about the holographic principle and came across the Planck area. It says that Planck area is the square of Planck length and there were some pictures like this:
source
I know that this Planck area is used for black holes. But doesn't matter if this Planck area is for a triangle or a square?
And there was a clip that Leonard Susskind was saying that Planck area is $10^{-33}$ cm on a side.
source 39:44.
But isn't it that the Planck length is $10^{-33}$ cm? If so, what is the "on a side"? Does it mean a face?
Please keep it simple so I can understand it.
| The Planck length is usually defined as
$$l_P=\sqrt\frac{\hbar G}{c^3}\approx 1.6\times 10^{-35}\,\text{m},$$
where $\hbar$ is the reduced Planck constant, $G$ is Newton’s gravitational constant, and $c$ is the speed of light.
The Planck area is usually defined as the square of the Planck length,
$$A_P=l_P^2=\frac{\hbar G}{c^3}\approx 2.6\times 10^{-70}\,\text{m}^2.$$
There are various quantum gravity theories, none yet accepted. The role of the Planck length and Planck area in physics is therefore still being researched. In particular, bits of information in black holes should not be pictured as being stored in nice little triangles or nice little squares on the event horizon. One bit of information probably “requires” about one Planck area, in some sense, but we don’t yet understand just how the information is stored so we can’t talk about it being in a particular place on the horizon or having a particular shape like a triangle or a square. That diagram is not to be taken literally.
As for the phrase “on a side”, it can he used to talk about the sizes of squares or triangles. A square has four sides and a triangle has three sides, so instead of talking about their area you can talk about the length of their sides.
| {
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Total force on upper block in two block system If a block $m$ is placed on another block $M$ and a force $F$ is applied on bolck $M$. Then how many forces are acting on block $m$.(Friction is non zero)
The image is taken from this site.
Is pseudo force acting on block $m$ or not?
| Three forces act on block $m$. (1) the force of gravity acting downward (2) the normal reaction force that block $M$ exerts on $m$ acting upward and equal to the downward force of gravity and (3) the friction force, if block $M$ accelerates, that block $M$ exerts on block $m$ horizontally in the same direction as as the external force applied to block $M$.
An example of a pseudo force is the apparent force you feel pressing you back against your car seat when you accelerate. It’s due to inertia and not some physical object pushing you against your seat. None of the three forces are pseudo forces
Hope this helps
| {
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Schwarzschild Radius of a Galaxy If an ultra compact/dense Galaxy has a Schwarzschild radius same as it is own radius, how can it be observed from the outside of the Galaxy?
|
If an ultra compact/dense Galaxy has a Schwarzschild radius same as it is own radius
This is not possible. The Schwarzschild radius of a galaxy with say 200.000 solar masses is 590 km. Disregarding the fact that all stars would reach the singularity in very short time (for comparison 13 hours in case of 3 billion solar masses) the theoretical matter density inside the black hole would be much much higher than in case these stars are packed next to each other.
| {
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